What is the general solution for cos(2x)+7sin(x)-4=0 ?

The general solution for cos(2x)+7sin(x)-4=0 is x= pi/6+2pin,x=(5pi)/6+2pin

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cos(2x)+7sin(x)-4=0

cos (2 x) + 7 sin (x) - 4 = 0

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

Solution steps

cos (2 x) + 7 sin (x) - 4 = 0

Rewrite using trig identities

- 3 - 2 sin^{2} (x) + 7 sin (x) = 0

Solve by substitution

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

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

sin (x) = 3 :

No Solution

Combine all the solutions

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

tan (\frac{θ}{8}) + 3 = 0

sin (\frac{θ}{2}) - 1 = 0, 0 \leq θ \leq 2 π

cos (2 x) = - \frac{2}{2}, 0 \leq x \leq 2 π

\frac{3 + sin ^{2} ( θ )}{cos ( θ ) - 2} = 3 cos (θ)

cos (a) = 0

What is the general solution for cos(2x)+7sin(x)-4=0 ?
The general solution for cos(2x)+7sin(x)-4=0 is x= pi/6+2pin,x=(5pi)/6+2pin