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

The general solution for 2cos^2(x)-sin(2x)-1=0 is x= pi/8+(pin)/2

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

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

x = \frac{π}{8} + \frac{πn}{2}

Solution steps

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

Rewrite using trig identities

1 - tan (2 x) = 0

Move

1

to the right side

- tan (2 x) = - 1

Divide both sides by

- 1

tan (2 x) = 1

General solutions for

tan (2 x) = 1

2 x = \frac{π}{4} + πn

Solve

2 x = \frac{π}{4} + πn : x = \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

sin (θ) = 0.84802194

x, sin (x + 6 0^{\circ}) = cos (y - 3 7^{\circ})

y, z = x^{2} sin (x y^{2})

tan (t) = \frac{4}{3}

cos (2 θ) = - sin^{2} (θ)

What is the general solution for 2cos^2(x)-sin(2x)-1=0 ?
The general solution for 2cos^2(x)-sin(2x)-1=0 is x= pi/8+(pin)/2