What is the general solution for 1-3sin^2(x)=cos^2(x)-sin^2(x) ?

The general solution for 1-3sin^2(x)=cos^2(x)-sin^2(x) is x=2pin,x=pi+2pin

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1-3sin^2(x)=cos^2(x)-sin^2(x)

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

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

Subtract

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

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

= - sin^{2} (x)

- sin^{2} (x) = 0

Divide both sides by

- 1

- sin^{2} (x) = 0

Divide both sides by

- 1

- sin^{2} (x) = 0

Divide both sides by

- 1

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

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

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

1 = l cos (x)

tan (2 x) = cos (2 x), 0 \leq x \leq π

5 sin (2 x) = 5 cos (x), 0 \leq x \leq 2 π

x, f = cos (x) cos (h y)

What is the general solution for 1-3sin^2(x)=cos^2(x)-sin^2(x) ?
The general solution for 1-3sin^2(x)=cos^2(x)-sin^2(x) is x=2pin,x=pi+2pin