What is the general solution for 3sin(2x)+1=2 ?

The general solution for 3sin(2x)+1=2 is x=(0.33983…)/2+pin,x= pi/2-(0.33983…)/2+pin

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3sin(2x)+1=2

3 sin (2 x) + 1 = 2

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

Degrees

x = 9.73561 \dots^{\circ} + 18 0^{\circ} n, x = 80.26438 \dots^{\circ} + 18 0^{\circ} n

Solution steps

3 sin (2 x) + 1 = 2

Move

1

to the right side

3 sin (2 x) + 1 = 2

Subtract

1

from both sides

3 sin (2 x) + 1 - 1 = 2 - 1

Simplify

3 sin (2 x) = 1

3 sin (2 x) = 1

Divide both sides by

3

3 sin (2 x) = 1

Divide both sides by

3

\frac{3 sin ( 2 x )}{3} = \frac{1}{3}

Simplify

sin (2 x) = \frac{1}{3}

sin (2 x) = \frac{1}{3}

Apply trig inverse properties

sin (2 x) = \frac{1}{3}

General solutions for

sin (2 x) = \frac{1}{3}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

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

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

Solve

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

2 x = arcsin (\frac{1}{3}) + 2 πn

Divide both sides by

2

2 x = arcsin (\frac{1}{3}) + 2 πn

Divide both sides by

2

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

Simplify

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

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

Solve

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

2 x = π - arcsin (\frac{1}{3}) + 2 πn

Divide both sides by

2

2 x = π - arcsin (\frac{1}{3}) + 2 πn

Divide both sides by

2

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

Simplify

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

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

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

Show solutions in decimal form

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

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x, t^{2} = 2 sec (x)

(8 cos (x) - 4) (sin (x)) = 0, 0 \leq x < 2 π

sin (x) = 3511

tan^{2} (x) - 1 = 4 cos (x)

What is the general solution for 3sin(2x)+1=2 ?
The general solution for 3sin(2x)+1=2 is x=(0.33983…)/2+pin,x= pi/2-(0.33983…)/2+pin