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

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

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

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

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

Solution steps

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

Subtract

2 sin^{2} (x)

from both sides

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

Rewrite using trig identities

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

Solve by substitution

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

cos (x) = 1 : x = 2 πn

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

Combine all the solutions

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

sec (θ) = \frac{7}{5}

\frac{sin ( x )}{cos ( x )} = 0

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

8 cos (x) - 42 = 0

3 csc (x) - 23 = 0

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