Solve 2cos2x = 7sinx?

Question

0<=x<=2pi

Answer 1

cos 2x = 1 - 2 sin^2 x

Letting u= sin x, you get:

2 (1-2u^2) = 7 u

Or:

4u^2 + 7u - 2 = 0

So u = [-7 +/- sqrt(49 - 4*4*(-2))]/8

49 - 4*4*(-2) = 49 + 32 = 81.

So:

u = [-7 +/- 9]/8 = 1/4 or u=-2.

Since sin x is between -1 and 1, u = sin x can't be -2.

So u = 1/4.

That means sin x = 1/4.

So x = arcsin(1/4) or x = pi - arcsin(1/4).

For approximations, that means:

x ~ 0.25269

or

x ~ 2.88891

Answer 2

2 (1 - 2 sin ² x) = 7 sin x
2 - 4 sin ² x = 7 sin x
4 sin ² x + 7sin x - 2 = 0
(4 sin x - 1)(sin x + 2) = 0
sin x = 1/4 is acceptable answer
x = 0.253 , x = 2.89 (in radians)