What is the general solution for cos^2(x)=1-sin(x),0<= x<2pi ?

The general solution for cos^2(x)=1-sin(x),0<= x<2pi is x=0,x=pi,x= pi/2

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cos^2(x)=1-sin(x),0<= x<2pi

cos^{2} (x) = 1 - sin (x), 0 \leq x < 2 π

x = 0, x = π, x = \frac{π}{2}

Solution steps

cos^{2} (x) = 1 - sin (x), 0 \leq x < 2 π

Subtract

1 - sin (x)

from both sides

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

Rewrite using trig identities

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

Solve by substitution

sin (x) = 0, sin (x) = 1

sin (x) = 0, 0 \leq x < 2 π : x = 0, x = π

sin (x) = 1, 0 \leq x < 2 π : x = \frac{π}{2}

Combine all the solutions

x = 0, x = π, x = \frac{π}{2}

cos (2 x) - 3 sin (x) = 2, 0^{\circ} \leq x \leq 36 0^{\circ}

2 cos^{2} (x) + 3 cos (x) = 4 cos (x) + 1

6 sin^{2} (x) + 5 sin (x) = 0

sin (x) = - \frac{1}{2}, (0, 2 π)

tan (θ) = \frac{4}{3}, cos (θ) < 0

What is the general solution for cos^2(x)=1-sin(x),0<= x<2pi ?
The general solution for cos^2(x)=1-sin(x),0<= x<2pi is x=0,x=pi,x= pi/2