What is the general solution for cos^2(x)-3sin^2(x)-1=0 ?

The general solution for cos^2(x)-3sin^2(x)-1=0 is x=2pin,x=pi+2pin

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cos^2(x)-3sin^2(x)-1=0

cos^{2} (x) - 3 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) - 3 sin^{2} (x) - 1 = 0

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

= - 4 sin^{2} (x)

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

Divide both sides by

- 4

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

Divide both sides by

- 4

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

Divide both sides by

- 4

\frac{- 4 sin ^{2} ( x )}{- 4} = \frac{0}{- 4}

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

tanh (x) + 4 sech (x) = 4

\frac{36}{49} + cos^{2} (θ) = 1

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

3 cos^{2} (x) + 10 cos (x) + 3 = 0

- 1 = sin (k) (45)

What is the general solution for cos^2(x)-3sin^2(x)-1=0 ?
The general solution for cos^2(x)-3sin^2(x)-1=0 is x=2pin,x=pi+2pin