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ArGM2015
We know that sin^2(x) = 1 - cos^2(x), so

2 - 2 cos^2(x) = 2 + cos(x)
cos(x) + 2 cos^2(x) = 0

One immediate possibility is that cos(x) = 0, in which case x = Pi/2 or 3pi/4. If cos(x) is not zero, then divide both sides by cos(x) to obtain

cos(x) = -1/2

which gives x = 2Pi/3 and x = 4Pi/3.

So your solutions are

Pi/2, 3Pi/4, 2Pi/3, and 4Pi/3

Hope this helps :D
olemakpadu
-sin²x = 2cosx - 2    .....................(a)

Note that:  sin²x + cos²x = 1

sin²x = 1 - cos²x,   substituting this into equation (a)

-sin²x = 2cosx - 2 

-(1 - cos²x) = 2cosx - 2 

-1 + cos²x = 2cosx - 2 

-1 + cos²x - 2cosx + 2 = 0

 cos²x - 2cosx + 2 - 1 = 0

 cos²x - 2cosx + 1 = 0

let p = cosx

 p² - 2p + 1 = 0       This is a quadratic equation.

 Let us factorize. The two factors are -p, and -p

 p² - 2p + 1 = 0 

 p² - p -p + 1 = 0 

p(p - 1) -1(p - 1) = 0

(p - 1)(p - 1) = 0

p - 1 = 0    or   p - 1 = 0

p = 0 + 1     or p = 0 + 1

p = 1  or 1 

p = 1 (twice)

Recall p = cosx

cosx = 1.             Recall Cosine is positive in the 1st and 4th quadrant. 

x = cos⁻¹(1)

x = 0°  and 360°   which is the same as  0 radians  or 2π radians  

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