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

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

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

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

x = π

Solution steps

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

Subtract

- sin^{2} (x) + 4

from both sides

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

Rewrite using trig identities

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

Solve by substitution

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

cos (x) = - 3, 0 \leq x \leq 2 π :

No Solution

cos (x) = - 1, 0 \leq x \leq 2 π : x = π

Combine all the solutions

x = π

0 = cos^{2} (\frac{π}{12} t)

sec^{2} (x) + 2 sec (x) + 4 = 0

cos (A) = 0.7865

sin (x) = cos (2 x + 5 1^{\circ})

sin (x) = \frac{π}{4}

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