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

The general solution for sin^2(x)=cos(x)-cos^2(x) is x=2pin

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

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

x = 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

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

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= - cos (x) + 1

- cos (x) + 1 = 0

Move

1

to the right side

- cos (x) + 1 = 0

Subtract

1

from both sides

- cos (x) + 1 - 1 = 0 - 1

Simplify

- cos (x) = - 1

- cos (x) = - 1

Divide both sides by

- 1

- cos (x) = - 1

Divide both sides by

- 1

\frac{- cos ( x )}{- 1} = \frac{- 1}{- 1}

Simplify

cos (x) = 1

cos (x) = 1

General solutions for

cos (x) = 1

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 = 0 + 2 πn

x = 0 + 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, arcsin (x + y) = 0

1 + cos (π x) = sin (π x)

sin (x - 1 5^{\circ}) = cos (2 0^{\circ} + x)

15 sin^{2} (x) - 17 sin (x) + 4 = 0

cos (x) = \frac{451.1}{497}

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