What is the general solution for 4sin(x)=-cos^2(x)+1,0<= x<= 2pi ?

The general solution for 4sin(x)=-cos^2(x)+1,0<= x<= 2pi is x=0,x=pi,x=2pi

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4sin(x)=-cos^2(x)+1,0<= x<= 2pi

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

x = 0, x = π, x = 2 π

Solution steps

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

Subtract

- cos^{2} (x) + 1

from both sides

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

Rewrite using trig identities

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

Solve by substitution

sin (x) = 0, sin (x) = 4

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

sin (x) = 4, 0 \leq x \leq 2 π :

No Solution

Combine all the solutions

x = 0, x = π, x = 2 π

3 tan (x) sin (x) - 2 tan (x) = 0

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

2 sin (θ) = - 3, 0 \leq θ < 2 π

cos^{2} (φ) = sin^{2} (φ)

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

What is the general solution for 4sin(x)=-cos^2(x)+1,0<= x<= 2pi ?
The general solution for 4sin(x)=-cos^2(x)+1,0<= x<= 2pi is x=0,x=pi,x=2pi