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

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

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

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

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

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

Subtract

1 + sin^{2} (x)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

= - 2 cos^{2} (x)

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

Divide both sides by

- 2

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

Divide both sides by

- 2

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

Divide both sides by

- 2

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

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

cos (θ) = \frac{11}{61}

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

cos (x) = \frac{4}{8}

\frac{sin ( 5 1 ^{\circ} )}{18} = \frac{sin ( x )}{22}

3 (1 + sin (θ)) = 3 (1 - sin (θ))

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