What is the general solution for 4sin(2x)-1=0 ?

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

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4sin(2x)-1=0

4 sin (2 x) - 1 = 0

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

Degrees

x = 7.23875 \dots^{\circ} + 18 0^{\circ} n, x = 82.76124 \dots^{\circ} + 18 0^{\circ} n

Solution steps

4 sin (2 x) - 1 = 0

Move

1

to the right side

4 sin (2 x) - 1 = 0

Add

1

to both sides

4 sin (2 x) - 1 + 1 = 0 + 1

Simplify

4 sin (2 x) = 1

4 sin (2 x) = 1

Divide both sides by

4

4 sin (2 x) = 1

Divide both sides by

4

\frac{4 sin ( 2 x )}{4} = \frac{1}{4}

Simplify

sin (2 x) = \frac{1}{4}

sin (2 x) = \frac{1}{4}

Apply trig inverse properties

sin (2 x) = \frac{1}{4}

General solutions for

sin (2 x) = \frac{1}{4}

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

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

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

Solve

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

2 x = arcsin (\frac{1}{4}) + 2 πn

Divide both sides by

2

2 x = arcsin (\frac{1}{4}) + 2 πn

Divide both sides by

2

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

Simplify

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

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

Solve

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

2 x = π - arcsin (\frac{1}{4}) + 2 πn

Divide both sides by

2

2 x = π - arcsin (\frac{1}{4}) + 2 πn

Divide both sides by

2

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

Simplify

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

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

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

Show solutions in decimal form

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

cos (2 x) - 3 sin (2 x) = 0

x, cos (x) + sin (x) = 0

sin (x) (2 cos (x) - 1) = 0

cos (θ) = 1 + 2 sec (θ)

sin (3 θ) = - 1, 0 \leq θ < 2 π

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