What is the general solution for 4sin(3x)-1=0 ?

The general solution for 4sin(3x)-1=0 is x=(0.25268…)/3+(2pin)/3 ,x= pi/3-(0.25268…)/3+(2pin)/3

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4sin(3x)-1=0

4 sin (3 x) - 1 = 0

x = \frac{0.25268 \dots}{3} + \frac{2 πn}{3}, x = \frac{π}{3} - \frac{0.25268 \dots}{3} + \frac{2 πn}{3}

Solution steps

4 sin (3 x) - 1 = 0

Move

1

to the right side

4 sin (3 x) = 1

Divide both sides by

4

sin (3 x) = \frac{1}{4}

Apply trig inverse properties

3 x = arcsin (\frac{1}{4}) + 2 πn, 3 x = π - arcsin (\frac{1}{4}) + 2 πn

Solve

3 x = arcsin (\frac{1}{4}) + 2 πn : x = \frac{arcsin ( \frac{1}{4} )}{3} + \frac{2 πn}{3}

Solve

3 x = π - arcsin (\frac{1}{4}) + 2 πn : x = \frac{π}{3} - \frac{arcsin ( \frac{1}{4} )}{3} + \frac{2 πn}{3}

x = \frac{arcsin ( \frac{1}{4} )}{3} + \frac{2 πn}{3}, x = \frac{π}{3} - \frac{arcsin ( \frac{1}{4} )}{3} + \frac{2 πn}{3}

Show solutions in decimal form

x = \frac{0.25268 \dots}{3} + \frac{2 πn}{3}, x = \frac{π}{3} - \frac{0.25268 \dots}{3} + \frac{2 πn}{3}

\frac{7}{4} cos (3 θ) - 1 = 0

cos (x) = sin (\frac{7 π}{6}), 0 \leq x \leq 2 π

cos (2 x) = \frac{( 3 k - 3 )}{8}

1.5 = 2.75 cos (30 (x - 4)) + 3.25

sin (x) = (\frac{5}{12})

What is the general solution for 4sin(3x)-1=0 ?
The general solution for 4sin(3x)-1=0 is x=(0.25268…)/3+(2pin)/3 ,x= pi/3-(0.25268…)/3+(2pin)/3