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

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

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

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

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

Solution steps

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

Subtract

sin^{2} (x)

from both sides

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

Rewrite using trig identities

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

Solve by substitution

cos (x) = 0, cos (x) = - 1

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = - 1 : x = π + 2 πn

Combine all the solutions

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

cos (x) = - \frac{3}{10}

tan (x - \frac{π}{2}) = \frac{1}{3}

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

3 tan (2 x) - 3 = 0

t, x = 2 sin (t)

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