What is the general solution for 6sin(3x)=2 ?

The general solution for 6sin(3x)=2 is x=(0.33983…)/3+(2pin)/3 ,x= pi/3-(0.33983…)/3+(2pin)/3

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6sin(3x)=2

6 sin (3 x) = 2

x = \frac{0.33983 \dots}{3} + \frac{2 πn}{3}, x = \frac{π}{3} - \frac{0.33983 \dots}{3} + \frac{2 πn}{3}

Solution steps

6 sin (3 x) = 2

Divide both sides by

6

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

Apply trig inverse properties

3 x = arcsin (\frac{1}{3}) + 2 πn, 3 x = π - arcsin (\frac{1}{3}) + 2 πn

Solve

3 x = arcsin (\frac{1}{3}) + 2 πn : x = \frac{arcsin ( \frac{1}{3} )}{3} + \frac{2 πn}{3}

Solve

3 x = π - arcsin (\frac{1}{3}) + 2 πn : x = \frac{π}{3} - \frac{arcsin ( \frac{1}{3} )}{3} + \frac{2 πn}{3}

x = \frac{arcsin ( \frac{1}{3} )}{3} + \frac{2 πn}{3}, x = \frac{π}{3} - \frac{arcsin ( \frac{1}{3} )}{3} + \frac{2 πn}{3}

Show solutions in decimal form

x = \frac{0.33983 \dots}{3} + \frac{2 πn}{3}, x = \frac{π}{3} - \frac{0.33983 \dots}{3} + \frac{2 πn}{3}

cos (x) \cdot sin (x) = 0

sin^{2} (θ) + 4 sin (θ) + 3 = 0

2 \cdot cos (x) = 0

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

- 2 sin^{2} (x) - 1 cos (x) = - 1

What is the general solution for 6sin(3x)=2 ?
The general solution for 6sin(3x)=2 is x=(0.33983…)/3+(2pin)/3 ,x= pi/3-(0.33983…)/3+(2pin)/3