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

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

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tan^2(x)-sin^2(x)=tan^2(x)+sin^2(x)

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

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

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

Subtract

tan^{2} (x) + sin^{2} (x)

from both sides

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

Divide both sides by

- 2

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

Divide both sides by

- 2

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

Divide both sides by

- 2

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

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

3 cos (2 x) + cos (x) = 0

sin (2 v) = sin (1 6^{\circ})

tan (2 x) = \frac{130}{- 270 - 120}

- 3 sin (x) - 2 cos (2 x) = 0

2 cos (x) tan (x) = 1

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