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

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

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

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

x = π

Solution steps

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

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

cos^{2} (θ) = 2 - 2 sin (θ)

x, sin (x) = 0.5

x, z = arcsin (\frac{y}{x})

3 csc (A) - 6 = 0

15 cos^{2} (x) + sin (x) - 9 = 0

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