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

The general solution for 4sin^2(x)=5-4cos(x),0<= x<2pi is x= pi/3 ,x=(5pi)/3

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

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

x = \frac{π}{3}, x = \frac{5 π}{3}

Solution steps

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

Subtract

5 - 4 cos (x)

from both sides

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

Rewrite using trig identities

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

Solve by substitution

cos (x) = \frac{1}{2}

cos (x) = \frac{1}{2}, 0 \leq x < 2 π : x = \frac{π}{3}, x = \frac{5 π}{3}

Combine all the solutions

x = \frac{π}{3}, x = \frac{5 π}{3}

sin (x) - 6 cos^{2} (x) + 4 = 0

cos (θ) = \frac{3}{4}

2 sec (4 x) - 2 = 0

A, y = - 6 cos (A) - 2 sin^{2} (A)

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

What is the general solution for 4sin^2(x)=5-4cos(x),0<= x<2pi ?
The general solution for 4sin^2(x)=5-4cos(x),0<= x<2pi is x= pi/3 ,x=(5pi)/3