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

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

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

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

x = \frac{π}{2} + 2 πn

Degrees

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

Solution steps

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

Subtract

sin^{2} (x)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= sin (x) - 1

sin (x) - 1 = 0

Move

1

to the right side

sin (x) - 1 = 0

Add

1

to both sides

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

Simplify

sin (x) = 1

sin (x) = 1

General solutions for

sin (x) = 1

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 = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

cos^{4} (x) = 1 + sin^{4} (x)

- 4 cos^{2} (θ) - 11 cos (θ) - 1 = - 7 cos (θ)

0 = 1 + cos (\frac{x}{2})

sin (x) = cos (1 9^{\circ})

\frac{1}{1.5} = \frac{sin ( 3 0 ^{\circ} )}{sin ( x )}, x < 9 0^{\circ}

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