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

The general solution for sin(x)=-cos^2(x)-1,0<= x<= 2pi is x=(3pi)/2

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

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

x = \frac{3 π}{2}

Solution steps

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

Subtract

- cos^{2} (x) - 1

from both sides

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

Rewrite using trig identities

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

Solve by substitution

sin (x) = - 1, sin (x) = 2

sin (x) = - 1, 0 \leq x \leq 2 π : x = \frac{3 π}{2}

sin (x) = 2, 0 \leq x \leq 2 π :

No Solution

Combine all the solutions

x = \frac{3 π}{2}

sin (x) (tan (x) - 3) = 0

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

cos (θ) tan (θ) + cot (θ) = csc (θ)

tan^{2} (x) = \frac{1}{cot ( x )}

\frac{sin ( θ )}{cos ( θ )} = 3

What is the general solution for sin(x)=-cos^2(x)-1,0<= x<= 2pi ?
The general solution for sin(x)=-cos^2(x)-1,0<= x<= 2pi is x=(3pi)/2