Solve a system of trigonometric equations












0














How can I solve this system of trigonometric equations analytically? It is from physics class.
$$
begin{cases}
30tcos{alpha}=50\
-30tsin{alpha}-4.9t^2=0
end{cases}
$$






trigonometry systems-of-equations nonlinear-system







share|cite|improve this question
























  • Hint: try moving the $t^2$ term to the other side, squaring the two equations, and adding them together. What happens to the trigonometric terms?
    – Josh B.
    Nov 28 '18 at 17:16


















0














How can I solve this system of trigonometric equations analytically? It is from physics class.
$$
begin{cases}
30tcos{alpha}=50\
-30tsin{alpha}-4.9t^2=0
end{cases}
$$






trigonometry systems-of-equations nonlinear-system







share|cite|improve this question
























  • Hint: try moving the $t^2$ term to the other side, squaring the two equations, and adding them together. What happens to the trigonometric terms?
    – Josh B.
    Nov 28 '18 at 17:16
















0












0








0


0





How can I solve this system of trigonometric equations analytically? It is from physics class.
$$
begin{cases}
30tcos{alpha}=50\
-30tsin{alpha}-4.9t^2=0
end{cases}
$$






trigonometry systems-of-equations nonlinear-system







share|cite|improve this question















How can I solve this system of trigonometric equations analytically? It is from physics class.
$$
begin{cases}
30tcos{alpha}=50\
-30tsin{alpha}-4.9t^2=0
end{cases}
$$






trigonometry systems-of-equations nonlinear-system




trigonometry systems-of-equations nonlinear-system






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share|cite|improve this question













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edited Nov 28 '18 at 17:12







Stepii

















asked Nov 28 '18 at 17:03









StepiiStepii

106




106












  • Hint: try moving the $t^2$ term to the other side, squaring the two equations, and adding them together. What happens to the trigonometric terms?
    – Josh B.
    Nov 28 '18 at 17:16




















  • Hint: try moving the $t^2$ term to the other side, squaring the two equations, and adding them together. What happens to the trigonometric terms?
    – Josh B.
    Nov 28 '18 at 17:16


















Hint: try moving the $t^2$ term to the other side, squaring the two equations, and adding them together. What happens to the trigonometric terms?
– Josh B.
Nov 28 '18 at 17:16






Hint: try moving the $t^2$ term to the other side, squaring the two equations, and adding them together. What happens to the trigonometric terms?
– Josh B.
Nov 28 '18 at 17:16












3 Answers
3






active

oldest

votes


















1














Hint: Squaring both the equations, you will get $900t^2cos^2{alpha}=2500\
900t^2sin^2{alpha}={4.9}^2t^4$.



Note that $sin^2{alpha}+cos^2{alpha}=1$.



So add both the equations and solve for $t$ using the substitution $t^2=u$.






share|cite|improve this answer

















  • 1




    Turn it into a quadratic is the gist of my answer but this is much nicer+faster.
    – Mason
    Nov 28 '18 at 17:22










  • @Mason Thank you.
    – Thomas Shelby
    Nov 28 '18 at 17:24



















1














$30tcos{alpha}=50 implies t=frac{5}{3} secalpha$



You can plug this information into the other equation and solve:



$$-30tsin{alpha}-4.9t^2=0implies -30(frac{5}{3} secalpha)sin{alpha}-4.9(frac{5}{3} secalpha)^2=0$$



$$-50(tanalpha)-4.9(frac{5}{3} secalpha)^2=0$$



$$-50(tanalpha)-4.9frac{25}{9} sec^2alpha=0$$



$$-50(tanalpha)-4.9frac{25}{9} (tan^2alpha+1)=0$$



Taking $y=tanalpha$ you can solve a quadratic equation.
$$-50(y)-4.9frac{25}{9} (y^2+1)=0$$



I think you're probably in good shape from here?






share|cite|improve this answer





















  • This results in negative alpha but its a real physics problem, so it must be positive.
    – Stepii
    Nov 28 '18 at 17:27










  • The angle cannot be negative? Add $2pi$?
    – Mason
    Nov 28 '18 at 17:29










  • Adding 2π gives angles greater than 90°
    – Stepii
    Nov 28 '18 at 17:38












  • @Stepii Which also doesn't make sense in the context of the problem?
    – Mason
    Nov 28 '18 at 17:43










  • What did you get for $y$. Maybe I made an error... Here's y
    – Mason
    Nov 28 '18 at 17:44





















0














Sorry. This is not an answer. Just part of a chat conversation that we are not really suppose to be having via comments.



enter image description here



This is the image from the link you shared. Unless I am reading this graph wrong it says: That when $a<0$ then $t>0$






share|cite|improve this answer





















  • The extraneous solutions are when $a>0$ because it implies that $t<0$ which as you have commented makes little sense often in physics problems.
    – Mason
    Nov 28 '18 at 18:23










  • But that solutions (with the angle and time both positive) satisfy the given system
    – Stepii
    Nov 28 '18 at 18:29











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














Hint: Squaring both the equations, you will get $900t^2cos^2{alpha}=2500\
900t^2sin^2{alpha}={4.9}^2t^4$.



Note that $sin^2{alpha}+cos^2{alpha}=1$.



So add both the equations and solve for $t$ using the substitution $t^2=u$.






share|cite|improve this answer

















  • 1




    Turn it into a quadratic is the gist of my answer but this is much nicer+faster.
    – Mason
    Nov 28 '18 at 17:22










  • @Mason Thank you.
    – Thomas Shelby
    Nov 28 '18 at 17:24
















1














Hint: Squaring both the equations, you will get $900t^2cos^2{alpha}=2500\
900t^2sin^2{alpha}={4.9}^2t^4$.



Note that $sin^2{alpha}+cos^2{alpha}=1$.



So add both the equations and solve for $t$ using the substitution $t^2=u$.






share|cite|improve this answer

















  • 1




    Turn it into a quadratic is the gist of my answer but this is much nicer+faster.
    – Mason
    Nov 28 '18 at 17:22










  • @Mason Thank you.
    – Thomas Shelby
    Nov 28 '18 at 17:24














1












1








1






Hint: Squaring both the equations, you will get $900t^2cos^2{alpha}=2500\
900t^2sin^2{alpha}={4.9}^2t^4$.



Note that $sin^2{alpha}+cos^2{alpha}=1$.



So add both the equations and solve for $t$ using the substitution $t^2=u$.






share|cite|improve this answer












Hint: Squaring both the equations, you will get $900t^2cos^2{alpha}=2500\
900t^2sin^2{alpha}={4.9}^2t^4$.



Note that $sin^2{alpha}+cos^2{alpha}=1$.



So add both the equations and solve for $t$ using the substitution $t^2=u$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 28 '18 at 17:18









Thomas ShelbyThomas Shelby

1,891217




1,891217








  • 1




    Turn it into a quadratic is the gist of my answer but this is much nicer+faster.
    – Mason
    Nov 28 '18 at 17:22










  • @Mason Thank you.
    – Thomas Shelby
    Nov 28 '18 at 17:24














  • 1




    Turn it into a quadratic is the gist of my answer but this is much nicer+faster.
    – Mason
    Nov 28 '18 at 17:22










  • @Mason Thank you.
    – Thomas Shelby
    Nov 28 '18 at 17:24








1




1




Turn it into a quadratic is the gist of my answer but this is much nicer+faster.
– Mason
Nov 28 '18 at 17:22




Turn it into a quadratic is the gist of my answer but this is much nicer+faster.
– Mason
Nov 28 '18 at 17:22












@Mason Thank you.
– Thomas Shelby
Nov 28 '18 at 17:24




@Mason Thank you.
– Thomas Shelby
Nov 28 '18 at 17:24











1














$30tcos{alpha}=50 implies t=frac{5}{3} secalpha$



You can plug this information into the other equation and solve:



$$-30tsin{alpha}-4.9t^2=0implies -30(frac{5}{3} secalpha)sin{alpha}-4.9(frac{5}{3} secalpha)^2=0$$



$$-50(tanalpha)-4.9(frac{5}{3} secalpha)^2=0$$



$$-50(tanalpha)-4.9frac{25}{9} sec^2alpha=0$$



$$-50(tanalpha)-4.9frac{25}{9} (tan^2alpha+1)=0$$



Taking $y=tanalpha$ you can solve a quadratic equation.
$$-50(y)-4.9frac{25}{9} (y^2+1)=0$$



I think you're probably in good shape from here?






share|cite|improve this answer





















  • This results in negative alpha but its a real physics problem, so it must be positive.
    – Stepii
    Nov 28 '18 at 17:27










  • The angle cannot be negative? Add $2pi$?
    – Mason
    Nov 28 '18 at 17:29










  • Adding 2π gives angles greater than 90°
    – Stepii
    Nov 28 '18 at 17:38












  • @Stepii Which also doesn't make sense in the context of the problem?
    – Mason
    Nov 28 '18 at 17:43










  • What did you get for $y$. Maybe I made an error... Here's y
    – Mason
    Nov 28 '18 at 17:44


















1














$30tcos{alpha}=50 implies t=frac{5}{3} secalpha$



You can plug this information into the other equation and solve:



$$-30tsin{alpha}-4.9t^2=0implies -30(frac{5}{3} secalpha)sin{alpha}-4.9(frac{5}{3} secalpha)^2=0$$



$$-50(tanalpha)-4.9(frac{5}{3} secalpha)^2=0$$



$$-50(tanalpha)-4.9frac{25}{9} sec^2alpha=0$$



$$-50(tanalpha)-4.9frac{25}{9} (tan^2alpha+1)=0$$



Taking $y=tanalpha$ you can solve a quadratic equation.
$$-50(y)-4.9frac{25}{9} (y^2+1)=0$$



I think you're probably in good shape from here?






share|cite|improve this answer





















  • This results in negative alpha but its a real physics problem, so it must be positive.
    – Stepii
    Nov 28 '18 at 17:27










  • The angle cannot be negative? Add $2pi$?
    – Mason
    Nov 28 '18 at 17:29










  • Adding 2π gives angles greater than 90°
    – Stepii
    Nov 28 '18 at 17:38












  • @Stepii Which also doesn't make sense in the context of the problem?
    – Mason
    Nov 28 '18 at 17:43










  • What did you get for $y$. Maybe I made an error... Here's y
    – Mason
    Nov 28 '18 at 17:44
















1












1








1






$30tcos{alpha}=50 implies t=frac{5}{3} secalpha$



You can plug this information into the other equation and solve:



$$-30tsin{alpha}-4.9t^2=0implies -30(frac{5}{3} secalpha)sin{alpha}-4.9(frac{5}{3} secalpha)^2=0$$



$$-50(tanalpha)-4.9(frac{5}{3} secalpha)^2=0$$



$$-50(tanalpha)-4.9frac{25}{9} sec^2alpha=0$$



$$-50(tanalpha)-4.9frac{25}{9} (tan^2alpha+1)=0$$



Taking $y=tanalpha$ you can solve a quadratic equation.
$$-50(y)-4.9frac{25}{9} (y^2+1)=0$$



I think you're probably in good shape from here?






share|cite|improve this answer












$30tcos{alpha}=50 implies t=frac{5}{3} secalpha$



You can plug this information into the other equation and solve:



$$-30tsin{alpha}-4.9t^2=0implies -30(frac{5}{3} secalpha)sin{alpha}-4.9(frac{5}{3} secalpha)^2=0$$



$$-50(tanalpha)-4.9(frac{5}{3} secalpha)^2=0$$



$$-50(tanalpha)-4.9frac{25}{9} sec^2alpha=0$$



$$-50(tanalpha)-4.9frac{25}{9} (tan^2alpha+1)=0$$



Taking $y=tanalpha$ you can solve a quadratic equation.
$$-50(y)-4.9frac{25}{9} (y^2+1)=0$$



I think you're probably in good shape from here?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 28 '18 at 17:20









MasonMason

1,9621530




1,9621530












  • This results in negative alpha but its a real physics problem, so it must be positive.
    – Stepii
    Nov 28 '18 at 17:27










  • The angle cannot be negative? Add $2pi$?
    – Mason
    Nov 28 '18 at 17:29










  • Adding 2π gives angles greater than 90°
    – Stepii
    Nov 28 '18 at 17:38












  • @Stepii Which also doesn't make sense in the context of the problem?
    – Mason
    Nov 28 '18 at 17:43










  • What did you get for $y$. Maybe I made an error... Here's y
    – Mason
    Nov 28 '18 at 17:44




















  • This results in negative alpha but its a real physics problem, so it must be positive.
    – Stepii
    Nov 28 '18 at 17:27










  • The angle cannot be negative? Add $2pi$?
    – Mason
    Nov 28 '18 at 17:29










  • Adding 2π gives angles greater than 90°
    – Stepii
    Nov 28 '18 at 17:38












  • @Stepii Which also doesn't make sense in the context of the problem?
    – Mason
    Nov 28 '18 at 17:43










  • What did you get for $y$. Maybe I made an error... Here's y
    – Mason
    Nov 28 '18 at 17:44


















This results in negative alpha but its a real physics problem, so it must be positive.
– Stepii
Nov 28 '18 at 17:27




This results in negative alpha but its a real physics problem, so it must be positive.
– Stepii
Nov 28 '18 at 17:27












The angle cannot be negative? Add $2pi$?
– Mason
Nov 28 '18 at 17:29




The angle cannot be negative? Add $2pi$?
– Mason
Nov 28 '18 at 17:29












Adding 2π gives angles greater than 90°
– Stepii
Nov 28 '18 at 17:38






Adding 2π gives angles greater than 90°
– Stepii
Nov 28 '18 at 17:38














@Stepii Which also doesn't make sense in the context of the problem?
– Mason
Nov 28 '18 at 17:43




@Stepii Which also doesn't make sense in the context of the problem?
– Mason
Nov 28 '18 at 17:43












What did you get for $y$. Maybe I made an error... Here's y
– Mason
Nov 28 '18 at 17:44






What did you get for $y$. Maybe I made an error... Here's y
– Mason
Nov 28 '18 at 17:44













0














Sorry. This is not an answer. Just part of a chat conversation that we are not really suppose to be having via comments.



enter image description here



This is the image from the link you shared. Unless I am reading this graph wrong it says: That when $a<0$ then $t>0$






share|cite|improve this answer





















  • The extraneous solutions are when $a>0$ because it implies that $t<0$ which as you have commented makes little sense often in physics problems.
    – Mason
    Nov 28 '18 at 18:23










  • But that solutions (with the angle and time both positive) satisfy the given system
    – Stepii
    Nov 28 '18 at 18:29
















0














Sorry. This is not an answer. Just part of a chat conversation that we are not really suppose to be having via comments.



enter image description here



This is the image from the link you shared. Unless I am reading this graph wrong it says: That when $a<0$ then $t>0$






share|cite|improve this answer





















  • The extraneous solutions are when $a>0$ because it implies that $t<0$ which as you have commented makes little sense often in physics problems.
    – Mason
    Nov 28 '18 at 18:23










  • But that solutions (with the angle and time both positive) satisfy the given system
    – Stepii
    Nov 28 '18 at 18:29














0












0








0






Sorry. This is not an answer. Just part of a chat conversation that we are not really suppose to be having via comments.



enter image description here



This is the image from the link you shared. Unless I am reading this graph wrong it says: That when $a<0$ then $t>0$






share|cite|improve this answer












Sorry. This is not an answer. Just part of a chat conversation that we are not really suppose to be having via comments.



enter image description here



This is the image from the link you shared. Unless I am reading this graph wrong it says: That when $a<0$ then $t>0$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 28 '18 at 18:22









MasonMason

1,9621530




1,9621530












  • The extraneous solutions are when $a>0$ because it implies that $t<0$ which as you have commented makes little sense often in physics problems.
    – Mason
    Nov 28 '18 at 18:23










  • But that solutions (with the angle and time both positive) satisfy the given system
    – Stepii
    Nov 28 '18 at 18:29


















  • The extraneous solutions are when $a>0$ because it implies that $t<0$ which as you have commented makes little sense often in physics problems.
    – Mason
    Nov 28 '18 at 18:23










  • But that solutions (with the angle and time both positive) satisfy the given system
    – Stepii
    Nov 28 '18 at 18:29
















The extraneous solutions are when $a>0$ because it implies that $t<0$ which as you have commented makes little sense often in physics problems.
– Mason
Nov 28 '18 at 18:23




The extraneous solutions are when $a>0$ because it implies that $t<0$ which as you have commented makes little sense often in physics problems.
– Mason
Nov 28 '18 at 18:23












But that solutions (with the angle and time both positive) satisfy the given system
– Stepii
Nov 28 '18 at 18:29




But that solutions (with the angle and time both positive) satisfy the given system
– Stepii
Nov 28 '18 at 18:29


















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