Solve on the interval $[0,2pi)$: $4 sin(x) cos(x)=1$. [closed]
I tried using the product-to-sum formulas, but did not come up with the correct answer.
algebra-precalculus trigonometry
closed as off-topic by Shailesh, amWhy, Toby Mak, Delta-u, José Carlos Santos Nov 29 '18 at 12:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, amWhy, Toby Mak, Delta-u, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
I tried using the product-to-sum formulas, but did not come up with the correct answer.
algebra-precalculus trigonometry
closed as off-topic by Shailesh, amWhy, Toby Mak, Delta-u, José Carlos Santos Nov 29 '18 at 12:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, amWhy, Toby Mak, Delta-u, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
I tried using the product-to-sum formulas, but did not come up with the correct answer.
algebra-precalculus trigonometry
I tried using the product-to-sum formulas, but did not come up with the correct answer.
algebra-precalculus trigonometry
algebra-precalculus trigonometry
edited Nov 29 '18 at 10:52
amWhy
192k28225439
192k28225439
asked Nov 29 '18 at 0:39
math818math818
33
33
closed as off-topic by Shailesh, amWhy, Toby Mak, Delta-u, José Carlos Santos Nov 29 '18 at 12:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, amWhy, Toby Mak, Delta-u, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Shailesh, amWhy, Toby Mak, Delta-u, José Carlos Santos Nov 29 '18 at 12:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, amWhy, Toby Mak, Delta-u, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
As was hinted, $$sin2x=2sin xcos x$$
Hence, your equation becomes
$$2sin2x=1$$
$$sin2x=frac12$$
$$2x=arcsinfrac12$$
$$2x=fracpi6,,frac{5pi}6$$
$$x=fracpi{12},,frac{5pi}{12}$$
add a comment |
Hint: use the double-angle formula for sine:
$$sin(2theta) = 2sin(theta)cos(theta)$$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
As was hinted, $$sin2x=2sin xcos x$$
Hence, your equation becomes
$$2sin2x=1$$
$$sin2x=frac12$$
$$2x=arcsinfrac12$$
$$2x=fracpi6,,frac{5pi}6$$
$$x=fracpi{12},,frac{5pi}{12}$$
add a comment |
As was hinted, $$sin2x=2sin xcos x$$
Hence, your equation becomes
$$2sin2x=1$$
$$sin2x=frac12$$
$$2x=arcsinfrac12$$
$$2x=fracpi6,,frac{5pi}6$$
$$x=fracpi{12},,frac{5pi}{12}$$
add a comment |
As was hinted, $$sin2x=2sin xcos x$$
Hence, your equation becomes
$$2sin2x=1$$
$$sin2x=frac12$$
$$2x=arcsinfrac12$$
$$2x=fracpi6,,frac{5pi}6$$
$$x=fracpi{12},,frac{5pi}{12}$$
As was hinted, $$sin2x=2sin xcos x$$
Hence, your equation becomes
$$2sin2x=1$$
$$sin2x=frac12$$
$$2x=arcsinfrac12$$
$$2x=fracpi6,,frac{5pi}6$$
$$x=fracpi{12},,frac{5pi}{12}$$
edited Nov 29 '18 at 1:55
answered Nov 29 '18 at 1:36
clathratusclathratus
3,325331
3,325331
add a comment |
add a comment |
Hint: use the double-angle formula for sine:
$$sin(2theta) = 2sin(theta)cos(theta)$$
add a comment |
Hint: use the double-angle formula for sine:
$$sin(2theta) = 2sin(theta)cos(theta)$$
add a comment |
Hint: use the double-angle formula for sine:
$$sin(2theta) = 2sin(theta)cos(theta)$$
Hint: use the double-angle formula for sine:
$$sin(2theta) = 2sin(theta)cos(theta)$$
edited Nov 29 '18 at 0:42
Bernard
118k639112
118k639112
answered Nov 29 '18 at 0:40
Eevee TrainerEevee Trainer
5,0471734
5,0471734
add a comment |
add a comment |