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

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

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

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

x = \frac{π}{2} + 2 πn

Degrees

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

Solution steps

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= 1

1 = sin (x)

Switch sides

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 (θ) + 2 sec (θ) = - 3

3 tan^{2} (θ) + 7 tan (θ) - 2 = 4

4 sin^{2} (x) = 0

5 cos (2 θ) = - 5 cos (θ)

sin (θ) = \frac{3}{5}, 0 < θ < \frac{π}{2}

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