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

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

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

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

x = \frac{π}{3}, x = \frac{5 π}{3}

Solution steps

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

Subtract

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

from both sides

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

Rewrite using trig identities

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

Solve by substitution

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

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

cos (x) = 2, 0 \leq x \leq 2 π :

No Solution

Combine all the solutions

x = \frac{π}{3}, x = \frac{5 π}{3}

4^{\frac{1}{3}} sec (x) = 4 (\frac{1}{4 ^{\frac{1}{3}}}) csc (x)

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

θ, tan (θ) = \frac{2 3}{2}

\frac{- 3 + 4 cos ^{2} ( θ )}{1 - 2 sin ( θ )} = a + b sin (θ)

2 sin^{2} (x) - 7 sin (x) = - 3, 0 \leq x \leq 2 π

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