How do you solve $cos 2 x = sin x$ over the interval 0 to 2pi?

Question

How do you solve $cos 2 x = sin x$ over the interval 0 to 2pi?

1 Answer

Impact of this question

1518 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-16T02:48:09

$pi/6, (5pi)/6 and (3pi)/2$

Explanation:

Apply the trig identity: $cos 2x = 1 - 2sin^2 x.$ We get:
$1 - 2sin^2 x = sin x$
$- 2cos^2 x - sin x + 1 = 0.$
Solve this quadratic equation for sin x.
Since a - b + c = 0, use shortcut. The 2 real roots are; -1 and
$-c/a = 1/2.$
Trig unit circle and trig table give -->
a. sin x = -1 --> $x = (3pi)/2.$
b. $sin x = 1/2$ --> $x = pi/6$ and $x = pi - pi/6 = (5pi)/6$

Science

Math

Humanities

... and beyond

How do you solve $cos 2 x = sin x$ over the interval 0 to 2pi?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve cos 2 x = sin xcos 2 x = sin x over the interval 0 to 2pi?

1 Answer

Explanation:

Related questions

Impact of this question

How do you solve $cos 2 x = sin x$ over the interval 0 to 2pi?