What is the general solution for 2sin^2(x)+9cos(x)-6=0 ?

The general solution for 2sin^2(x)+9cos(x)-6=0 is x= pi/3+2pin,x=(5pi)/3+2pin

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2sin^2(x)+9cos(x)-6=0

2 sin^{2} (x) + 9 cos (x) - 6 = 0

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

Solution steps

2 sin^{2} (x) + 9 cos (x) - 6 = 0

Rewrite using trig identities

- 4 - 2 cos^{2} (x) + 9 cos (x) = 0

Solve by substitution

cos (x) = \frac{1}{2}, cos (x) = 4

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

cos (x) = 4 :

No Solution

Combine all the solutions

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

tan (θ) = \frac{- 4}{- 3}

cos^{2} (x) + 3 cos (x) + 2 = 0

5 tan^{2} (x) + 5 = 10

sin (6 x) + sin (4 x) + sin (2 x) = 0

y, x = sin (2 y)

What is the general solution for 2sin^2(x)+9cos(x)-6=0 ?
The general solution for 2sin^2(x)+9cos(x)-6=0 is x= pi/3+2pin,x=(5pi)/3+2pin