What is the general solution for 4cos^2(x)=5+4sin(x),0<= x<= 2pi ?

The general solution for 4cos^2(x)=5+4sin(x),0<= x<= 2pi is x=(7pi)/6 ,x=(11pi)/6

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4cos^2(x)=5+4sin(x),0<= x<= 2pi

4 cos^{2} (x) = 5 + 4 sin (x), 0 \leq x \leq 2 π

x = \frac{7 π}{6}, x = \frac{11 π}{6}

Solution steps

4 cos^{2} (x) = 5 + 4 sin (x), 0 \leq x \leq 2 π

Subtract

5 + 4 sin (x)

from both sides

4 cos^{2} (x) - 5 - 4 sin (x) = 0

Rewrite using trig identities

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

Solve by substitution

sin (x) = - \frac{1}{2}

sin (x) = - \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{7 π}{6}, x = \frac{11 π}{6}

Combine all the solutions

x = \frac{7 π}{6}, x = \frac{11 π}{6}

sin (π + θ) = sin (θ)

0 = 2 cos (x) + sin (x)

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

tan (t) + sec (t) = 1

sin (5 x - 10) = cos (x - 8)

What is the general solution for 4cos^2(x)=5+4sin(x),0<= x<= 2pi ?
The general solution for 4cos^2(x)=5+4sin(x),0<= x<= 2pi is x=(7pi)/6 ,x=(11pi)/6