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

The general solution for 5sin(3x)=-2 is x=-(0.41151…)/3+(2pin)/3 ,x= pi/3+(0.41151…)/3+(2pin)/3

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5sin(3x)=-2

5 sin (3 x) = - 2

x = - \frac{0.41151 \dots}{3} + \frac{2 πn}{3}, x = \frac{π}{3} + \frac{0.41151 \dots}{3} + \frac{2 πn}{3}

Solution steps

5 sin (3 x) = - 2

Divide both sides by

5

sin (3 x) = - \frac{2}{5}

Apply trig inverse properties

3 x = arcsin (- \frac{2}{5}) + 2 πn, 3 x = π + arcsin (\frac{2}{5}) + 2 πn

Solve

3 x = arcsin (- \frac{2}{5}) + 2 πn : x = - \frac{arcsin ( \frac{2}{5} )}{3} + \frac{2 πn}{3}

Solve

3 x = π + arcsin (\frac{2}{5}) + 2 πn : x = \frac{π}{3} + \frac{arcsin ( \frac{2}{5} )}{3} + \frac{2 πn}{3}

x = - \frac{arcsin ( \frac{2}{5} )}{3} + \frac{2 πn}{3}, x = \frac{π}{3} + \frac{arcsin ( \frac{2}{5} )}{3} + \frac{2 πn}{3}

Show solutions in decimal form

x = - \frac{0.41151 \dots}{3} + \frac{2 πn}{3}, x = \frac{π}{3} + \frac{0.41151 \dots}{3} + \frac{2 πn}{3}

cos^{2} (x) - 1 = 0, 0 \leq x < 2 π

tan^{2} (x) + 1 = 2 \cdot tan (x)

\frac{csc ( θ )}{cot ( θ )} = - 2 tan (θ)

2 cos^{2} (θ) + 3 sin (θ) = 0

sin (2 θ) + 3 cos (θ) = 0

What is the general solution for 5sin(3x)=-2 ?
The general solution for 5sin(3x)=-2 is x=-(0.41151…)/3+(2pin)/3 ,x= pi/3+(0.41151…)/3+(2pin)/3