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

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

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

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

x = 2 πn, x = π + 2 πn

Solution steps

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

Rewrite using trig identities

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

Divide both sides by

- 3

sin (x) = 0

General solutions for

sin (x) = 0

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 2 πn, x = π + 2 πn

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

4 sin^{2} (x) + 4 cos (2 x) = 1

100 + 60 sin (\frac{7 π}{3} x) = 100, 0 \leq x \leq 1

cot (θ) = 4, cos (θ) < 0

(tan (x) + 1) (sin (x) - 1) = 0

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