What is the general solution for 7sin(2x)+12sin(x)=0 ?

The general solution for 7sin(2x)+12sin(x)=0 is x=2pin,x=pi+2pin,x=2.60049…+2pin,x=-2.60049…+2pin

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7sin(2x)+12sin(x)=0

7 sin (2 x) + 12 sin (x) = 0

x = 2 πn, x = π + 2 πn, x = 2.60049 \dots + 2 πn, x = - 2.60049 \dots + 2 πn

Solution steps

7 sin (2 x) + 12 sin (x) = 0

Rewrite using trig identities

12 sin (x) + 14 cos (x) sin (x) = 0

Factor

12 sin (x) + 14 cos (x) sin (x) : 2 sin (x) (7 cos (x) + 6)

2 sin (x) (7 cos (x) + 6) = 0

Solving each part separately

sin (x) = 0 or 7 cos (x) + 6 = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

7 cos (x) + 6 = 0 : x = arccos (- \frac{6}{7}) + 2 πn, x = - arccos (- \frac{6}{7}) + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = arccos (- \frac{6}{7}) + 2 πn, x = - arccos (- \frac{6}{7}) + 2 πn

Show solutions in decimal form

x = 2 πn, x = π + 2 πn, x = 2.60049 \dots + 2 πn, x = - 2.60049 \dots + 2 πn

x, cos (x) = cos (2 x)

cos (x) = 0.50074242

2 sin (4 x) - 1 = 0, - 2 π \leq x \leq 3 π

tan (θ) = \frac{10}{10}

0 = sin (\frac{θ}{2}) - sin (θ)

What is the general solution for 7sin(2x)+12sin(x)=0 ?
The general solution for 7sin(2x)+12sin(x)=0 is x=2pin,x=pi+2pin,x=2.60049…+2pin,x=-2.60049…+2pin