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

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

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

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

x = πn

Solution steps

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

Rewrite using trig identities

- 1 + cos (2 x) = 0

Move

1

to the right side

cos (2 x) = 1

General solutions for

cos (2 x) = 1

2 x = 0 + 2 πn

Solve

2 x = 0 + 2 πn : x = πn

x = πn

cos^{2} (x) + 6 cos (x) + 5 = 0

sin (t) = - 0.9397

tan^{2} (x) = 2 sec^{2} (x) - 3

sinh (x) + 4 = 4 cosh (x)

sin (18 x) = - 1

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