What is the general solution for cos^2(x)+2cos(x)=sin^2(x/2)-2 ?

The general solution for cos^2(x)+2cos(x)=sin^2(x/2)-2 is x=pi+2pin

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

cos^{2} (x) + 2 cos (x) = sin^{2} (\frac{x}{2}) - 2

x = π + 2 πn

Solution steps

cos^{2} (x) + 2 cos (x) = sin^{2} (\frac{x}{2}) - 2

Subtract

sin^{2} (\frac{x}{2}) - 2

from both sides

cos^{2} (x) + 2 cos (x) - sin^{2} (\frac{x}{2}) + 2 = 0

Rewrite using trig identities

2 + \frac{- 1 + 2 cos ^{2} ( x ) + 5 cos ( x )}{2} = 0

Solve by substitution

cos (x) = - 1, cos (x) = - \frac{3}{2}

cos (x) = - 1 : x = π + 2 πn

cos (x) = - \frac{3}{2} :

No Solution

Combine all the solutions

x = π + 2 πn

2 cos^{2} (x) - 4 cos (x) + 2 = 0

cos^{2} (x) - sin^{2} (x) = 1 - 2 sin (x)

cos (x) = - 0.34

2 sin^{2} (x) = 3 sin (x) - 1

2 cos (x) sin (x) - sin (x) = 0

What is the general solution for cos^2(x)+2cos(x)=sin^2(x/2)-2 ?
The general solution for cos^2(x)+2cos(x)=sin^2(x/2)-2 is x=pi+2pin