What is the general solution for 5sin(2x)+9sin(x)=0 ?

The general solution for 5sin(2x)+9sin(x)=0 is x=2pin,x=pi+2pin,x=2.69056…+2pin,x=-2.69056…+2pin

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5sin(2x)+9sin(x)=0

5 sin (2 x) + 9 sin (x) = 0

x = 2 πn, x = π + 2 πn, x = 2.69056 \dots + 2 πn, x = - 2.69056 \dots + 2 πn

Solution steps

5 sin (2 x) + 9 sin (x) = 0

Rewrite using trig identities

9 sin (x) + 10 cos (x) sin (x) = 0

Factor

9 sin (x) + 10 cos (x) sin (x) : sin (x) (10 cos (x) + 9)

sin (x) (10 cos (x) + 9) = 0

Solving each part separately

sin (x) = 0 or 10 cos (x) + 9 = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

10 cos (x) + 9 = 0 : x = arccos (- \frac{9}{10}) + 2 πn, x = - arccos (- \frac{9}{10}) + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = arccos (- \frac{9}{10}) + 2 πn, x = - arccos (- \frac{9}{10}) + 2 πn

Show solutions in decimal form

x = 2 πn, x = π + 2 πn, x = 2.69056 \dots + 2 πn, x = - 2.69056 \dots + 2 πn

4 sin (x) - 6 cos (x) = 3

4 cos (x) = cos^{3} (x)

cos^{2} (a) = \frac{3}{5 sin ( a )}

3 cos (x) = 2 sec (x) - 5

sin (\frac{a}{2}) = (\frac{1}{2}) sin (a)

What is the general solution for 5sin(2x)+9sin(x)=0 ?
The general solution for 5sin(2x)+9sin(x)=0 is x=2pin,x=pi+2pin,x=2.69056…+2pin,x=-2.69056…+2pin