What is the general solution for 2cos^2(x)+15sin(x)=9 ?

The general solution for 2cos^2(x)+15sin(x)=9 is x= pi/6+2pin,x=(5pi)/6+2pin

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2cos^2(x)+15sin(x)=9

2 cos^{2} (x) + 15 sin (x) = 9

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

Solution steps

2 cos^{2} (x) + 15 sin (x) = 9

Subtract

9

from both sides

2 cos^{2} (x) + 15 sin (x) - 9 = 0

Rewrite using trig identities

- 7 + 15 sin (x) - 2 sin^{2} (x) = 0

Solve by substitution

sin (x) = \frac{1}{2}, sin (x) = 7

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

sin (x) = 7 :

No Solution

Combine all the solutions

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

cot (θ) = 1.413

arccos (x) + arccos (2 x) = arccos (\frac{1}{2})

tan (θ) = \frac{2 3}{3} sin (θ), 0 \leq θ \leq 2 π

x, cos (x + y) = - cos (x)

4 + 2 tan^{2} (x) = 5 tan^{2} (x) + 3

What is the general solution for 2cos^2(x)+15sin(x)=9 ?
The general solution for 2cos^2(x)+15sin(x)=9 is x= pi/6+2pin,x=(5pi)/6+2pin