What is the general solution for sin^2(θ)+2cos(θ)= 7/4 ?

The general solution for sin^2(θ)+2cos(θ)= 7/4 is θ= pi/3+2pin,θ=(5pi)/3+2pin

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sin^2(θ)+2cos(θ)= 7/4

sin^{2} (θ) + 2 cos (θ) = \frac{7}{4}

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Solution steps

sin^{2} (θ) + 2 cos (θ) = \frac{7}{4}

Subtract

\frac{7}{4}

from both sides

sin^{2} (θ) + 2 cos (θ) - \frac{7}{4} = 0

Simplify

sin^{2} (θ) + 2 cos (θ) - \frac{7}{4} : \frac{4 sin ^{2} ( θ ) + 8 cos ( θ ) - 7}{4}

\frac{4 sin ^{2} ( θ ) + 8 cos ( θ ) - 7}{4} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

4 sin^{2} (θ) + 8 cos (θ) - 7 = 0

Rewrite using trig identities

- 3 - 4 cos^{2} (θ) + 8 cos (θ) = 0

Solve by substitution

cos (θ) = \frac{1}{2}, cos (θ) = \frac{3}{2}

cos (θ) = \frac{1}{2} : θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

cos (θ) = \frac{3}{2} :

No Solution

Combine all the solutions

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

5 cos (x) + 3 = 3 cos (x) + 5

sin (\frac{π}{2}) = cos (x)

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

x, arcsin (\frac{x}{16}) = 31

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

What is the general solution for sin^2(θ)+2cos(θ)= 7/4 ?
The general solution for sin^2(θ)+2cos(θ)= 7/4 is θ= pi/3+2pin,θ=(5pi)/3+2pin