What is the general solution for 5sin(3x)-1=3 ?

The general solution for 5sin(3x)-1=3 is x=(0.92729…)/3+(2pin)/3 ,x= pi/3-(0.92729…)/3+(2pin)/3

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5sin(3x)-1=3

5 sin (3 x) - 1 = 3

x = \frac{0.92729 \dots}{3} + \frac{2 πn}{3}, x = \frac{π}{3} - \frac{0.92729 \dots}{3} + \frac{2 πn}{3}

Solution steps

5 sin (3 x) - 1 = 3

Move

1

to the right side

5 sin (3 x) = 4

Divide both sides by

5

sin (3 x) = \frac{4}{5}

Apply trig inverse properties

3 x = arcsin (\frac{4}{5}) + 2 πn, 3 x = π - arcsin (\frac{4}{5}) + 2 πn

Solve

3 x = arcsin (\frac{4}{5}) + 2 πn : x = \frac{arcsin ( \frac{4}{5} )}{3} + \frac{2 πn}{3}

Solve

3 x = π - arcsin (\frac{4}{5}) + 2 πn : x = \frac{π}{3} - \frac{arcsin ( \frac{4}{5} )}{3} + \frac{2 πn}{3}

x = \frac{arcsin ( \frac{4}{5} )}{3} + \frac{2 πn}{3}, x = \frac{π}{3} - \frac{arcsin ( \frac{4}{5} )}{3} + \frac{2 πn}{3}

Show solutions in decimal form

x = \frac{0.92729 \dots}{3} + \frac{2 πn}{3}, x = \frac{π}{3} - \frac{0.92729 \dots}{3} + \frac{2 πn}{3}

- 5 sin (x) = 0

sin^{2} (θ) = 6 (cos (- θ) + 1)

5 sin (θ) - 2 = 0

(2 cos (x) - 2) (2 sin (x) - 2) = 0

- 4 cos (5 x) + 1 = 1

What is the general solution for 5sin(3x)-1=3 ?
The general solution for 5sin(3x)-1=3 is x=(0.92729…)/3+(2pin)/3 ,x= pi/3-(0.92729…)/3+(2pin)/3