How do I isolate x in 2cos2x - sinx = -2/pi?

cos2x = 1 - sin^2(x)

Therefore,

2-2sin^2(x) - sinx =-2/pi

2+2/pi = 2sin^2(x)+sinx

thus ,

2sin^2(x)+sinx = (2+2/pi)

2sin^2(x)+sinx - (2+2/pi) = 0

use the quadratic formula, where a = 2 b=1 c= -(2+2/pi)

In addition, there are 3 main indentities for cos2x !!!

cos(2x) = cos^2(x) - sin^2(x) [double angle formula]

2 cos^2(x) - 2 sin^2(x) - sin(x) = -2/Pi

Now cos^2(x) = 1 - sin^2(x)

2 - 2 sin^2(x) - 2 sin^2(x) - sin(x) = -2/Pi

Let u = sin(x)

-4u^2 - u + ( 2 + 2/Pi ) = 0

By the quadratic formula

u = -1/8 (+/-) sqrt[ 1 + 8*( 2 + 2/Pi ) ] / 8

u = -1/8 (+/-) sqrt[ 17 + 16 / Pi ] / 8

Resubstitute for u

u = sin(x) = -1/8 (+/-) sqrt[ 17 + 16 / Pi ] / 8

x = sin^(-1)[ -1/8 (+/-) sqrt[ 17 + 16 / Pi ] / 8 ]

I'll let you do the arithmetic.

I hope this helps