What is the general solution for 1+cos(x)=2-2sin^2(x) ?

The general solution for 1+cos(x)=2-2sin^2(x) is x=(2pi)/3+2pin,x=(4pi)/3+2pin,x=2pin

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

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

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = 2 πn

Solution steps

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

Subtract

2 - 2 sin^{2} (x)

from both sides

2 sin^{2} (x) + cos (x) - 1 = 0

Rewrite using trig identities

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

Solve by substitution

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

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

cos (x) = 1 : x = 2 πn

Combine all the solutions

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = 2 πn

sin (θ - \frac{π}{4}) + 1 = 0

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

cos^{2} (θ) - cos (θ) - 12 = 0

(sin (3 x) - 3) (cos (x) + 1) = 0

sin (c) = \frac{2}{π ( - cos ( c ) + 1 )}

What is the general solution for 1+cos(x)=2-2sin^2(x) ?
The general solution for 1+cos(x)=2-2sin^2(x) is x=(2pi)/3+2pin,x=(4pi)/3+2pin,x=2pin