What is the general solution for 2cos^2(θ)+7sin(θ)=5 ?

The general solution for 2cos^2(θ)+7sin(θ)=5 is θ= pi/6+2pin,θ=(5pi)/6+2pin

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2cos^2(θ)+7sin(θ)=5

2 cos^{2} (θ) + 7 sin (θ) = 5

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Solution steps

2 cos^{2} (θ) + 7 sin (θ) = 5

Subtract

5

from both sides

2 cos^{2} (θ) + 7 sin (θ) - 5 = 0

Rewrite using trig identities

- 3 - 2 sin^{2} (θ) + 7 sin (θ) = 0

Solve by substitution

sin (θ) = \frac{1}{2}, sin (θ) = 3

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sin (θ) = 3 :

No Solution

Combine all the solutions

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

\frac{2 3}{3} = sec (θ)

4 (sin (x))^{2} - 15 sin (x) - 4 = 0

1 + sin (θ) = 1 + sin (- θ)

cos (2 x) = 2 + 3 sin (x), \frac{π}{2} \leq x \leq \frac{3 π}{2}

tan (θ) = (\frac{3}{8})

What is the general solution for 2cos^2(θ)+7sin(θ)=5 ?
The general solution for 2cos^2(θ)+7sin(θ)=5 is θ= pi/6+2pin,θ=(5pi)/6+2pin