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

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

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

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

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

Solution steps

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

Subtract

2 sin^{2} (x)

from both sides

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

Rewrite using trig identities

2 cos (2 x) = 0

Divide both sides by

2

cos (2 x) = 0

General solutions for

cos (2 x) = 0

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

Solve

2 x = \frac{π}{2} + 2 πn : x = \frac{π}{4} + πn

Solve

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

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

6 sin^{2} (\frac{θ}{2}) = cos^{2} (θ) + 5

cos (x) = \frac{1}{4} - sin^{2} (x)

2 sin^{2} (x) - 3 sin (x) cos (x) + cos^{2} (x) = 0

3 cos^{2} (x) = sin (x)

sin (b) = \frac{3}{5}

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