What is the general solution for -4sin(x)=-cos^2(x)+4,0<= x<2pi ?

The general solution for -4sin(x)=-cos^2(x)+4,0<= x<2pi is x=(3pi)/2

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-4sin(x)=-cos^2(x)+4,0<= x<2pi

- 4 sin (x) = - cos^{2} (x) + 4, 0 \leq x < 2 π

x = \frac{3 π}{2}

Solution steps

- 4 sin (x) = - cos^{2} (x) + 4, 0 \leq x < 2 π

Subtract

- cos^{2} (x) + 4

from both sides

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

Rewrite using trig identities

- 3 - sin^{2} (x) - 4 sin (x) = 0

Solve by substitution

sin (x) = - 3, sin (x) = - 1

sin (x) = - 3, 0 \leq x < 2 π :

No Solution

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

Combine all the solutions

x = \frac{3 π}{2}

cos (x) = \frac{3}{2}

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

sin (x) = 7938894612

3 sin (2 θ) = 5 cos (θ)

cos (2 x) = \frac{1}{2}, (0, 2 π)

What is the general solution for -4sin(x)=-cos^2(x)+4,0<= x<2pi ?
The general solution for -4sin(x)=-cos^2(x)+4,0<= x<2pi is x=(3pi)/2