What is the general solution for 0=cos(x)(1-cos(x))+sin^2(x) ?

The general solution for 0=cos(x)(1-cos(x))+sin^2(x) is x=(2pi)/3+2pin,x=(4pi)/3+2pin,x=2pin

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

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

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

Solution steps

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

Switch sides

cos (x) (1 - cos (x)) + 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{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 (2 x) = 3 cos^{2} (x)

\frac{1}{cos ( x )} = 1.98

2 sin^{2} (3 θ) - 1 = 0

4 sin (4 θ - \frac{π}{3}) + 3 = 5

2 tan (x) - 3 = 0

What is the general solution for 0=cos(x)(1-cos(x))+sin^2(x) ?
The general solution for 0=cos(x)(1-cos(x))+sin^2(x) is x=(2pi)/3+2pin,x=(4pi)/3+2pin,x=2pin