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

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Simplify

- 1 + 1 - 2 sin^{2} (x) - sin^{2} (x) : - 3 sin^{2} (x)

- 1 + 1 - 2 sin^{2} (x) - sin^{2} (x)

Add similar elements:

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

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

- 1 + 1 = 0

= - 3 sin^{2} (x)

= - 3 sin^{2} (x)

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

Divide both sides by

- 3

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

Divide both sides by

- 3

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

Divide both sides by

- 3

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

Simplify

sin^{2} (x) = 0

sin^{2} (x) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

sin (x) = 0

General solutions for

sin (x) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 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