What is the general solution for 3cos(2x)-5sin(x)+1=0 ?

The general solution for 3cos(2x)-5sin(x)+1=0 is x= pi/6+2pin,x=(5pi)/6+2pin

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3cos(2x)-5sin(x)+1=0

3 cos (2 x) - 5 sin (x) + 1 = 0

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

Solution steps

3 cos (2 x) - 5 sin (x) + 1 = 0

Rewrite using trig identities

4 - 5 sin (x) - 6 sin^{2} (x) = 0

Solve by substitution

sin (x) = - \frac{4}{3}, sin (x) = \frac{1}{2}

sin (x) = - \frac{4}{3} :

No Solution

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

Combine all the solutions

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

tan (θ) = \frac{5}{- 2}

sin (x + \frac{π}{2}) = \frac{1}{2}

8 cos (x) + 4 = 0

6 = 6 (1 - sin (θ))

tan (x) = 3, - π \leq x < π

What is the general solution for 3cos(2x)-5sin(x)+1=0 ?
The general solution for 3cos(2x)-5sin(x)+1=0 is x= pi/6+2pin,x=(5pi)/6+2pin