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

The general solution for 1+cos(x)=2sin^2(x) is x= pi/3+2pin,x=(5pi)/3+2pin,x=pi+2pin

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

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

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = π + 2 πn

Solution steps

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

Subtract

2 sin^{2} (x)

from both sides

1 + cos (x) - 2 sin^{2} (x) = 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{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = - 1 : x = π + 2 πn

Combine all the solutions

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = π + 2 πn

cos (x) = \frac{7}{25}

2 sin (3 x) + 2 = 0

0 = cos (θ)

sec^{2} (x) + 3 tan (x) = 5

2 sin (θ) + 1 = csc (θ)

What is the general solution for 1+cos(x)=2sin^2(x) ?
The general solution for 1+cos(x)=2sin^2(x) is x= pi/3+2pin,x=(5pi)/3+2pin,x=pi+2pin