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

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

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

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

x = 2 πn, x = π + 2 πn

Solution steps

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

Subtract

1

from both sides

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

Rewrite using trig identities

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

Solve by substitution

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

cos (x) = 1 : x = 2 πn

cos (x) = - 1 : x = π + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn

4 tan (3 x) = - 4

2 sin^{2} (x) - 1 + cos (x) = 0, 0 \leq x \leq 2 π

x, arctan (x) = 4 t + c

cos (\frac{π}{3} - x) = 1

2 (cos (t))^{2} - cos (t) - 1 = 0

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