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

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

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

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

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

1

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

= - 3 cos^{2} (x)

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

Divide both sides by

- 3

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

Divide both sides by

- 3

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

Divide both sides by

- 3

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

Simplify

cos^{2} (x) = 0

cos^{2} (x) = 0

Apply rule

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

cos (x) = 0

General solutions for

cos (x) = 0

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

sin (x) = - \frac{12}{13}

2 sin^{2} (x) + 7 sin (x) - 4 = 0

cos (\frac{5 π}{4} + x) + sin (\frac{5 π}{4} - x) = 0

cos (2 x) + sin (2 x) = 1

θ, sin (θ) = \frac{1}{2}

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