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

The general solution for (sin(2x))/(4sin(2x)-1)=1 is x=(0.33983…)/2+pin,x= pi/2-(0.33983…)/2+pin

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Popular Trigonometry >

(sin(2x))/(4sin(2x)-1)=1

Solution

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

Solution

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

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

Solve by substitution

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

Let:

sin (2 x) = u

\frac{u}{4 u - 1} = 1

\frac{u}{4 u - 1} = 1 : u = \frac{1}{3}

\frac{u}{4 u - 1} = 1

Multiply both sides by

4 u - 1

\frac{u}{4 u - 1} = 1

Multiply both sides by

4 u - 1

\frac{u}{4 u - 1} (4 u - 1) = 1 \cdot (4 u - 1)

Simplify

\frac{u}{4 u - 1} (4 u - 1) = 1 \cdot (4 u - 1)

Simplify

\frac{u}{4 u - 1} (4 u - 1) : u

\frac{u}{4 u - 1} (4 u - 1)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{u ( 4 u - 1 )}{4 u - 1}

Cancel the common factor:

4 u - 1

= u

Simplify

1 \cdot (4 u - 1) : 4 u - 1

1 \cdot (4 u - 1)

Multiply:

1 \cdot (4 u - 1) = (4 u - 1)

= (4 u - 1)

Remove parentheses:

(a) = a

= 4 u - 1

u = 4 u - 1

u = 4 u - 1

u = 4 u - 1

Move

4 u

to the left side

u = 4 u - 1

Subtract

4 u

from both sides

u - 4 u = 4 u - 1 - 4 u

Simplify

- 3 u = - 1

- 3 u = - 1

Divide both sides by

- 3

- 3 u = - 1

Divide both sides by

- 3

\frac{- 3 u}{- 3} = \frac{- 1}{- 3}

Simplify

u = \frac{1}{3}

u = \frac{1}{3}

Verify Solutions

Find undefined (singularity) points:

u = \frac{1}{4}

Take the denominator(s) of

\frac{u}{4 u - 1}

and compare to zero

Solve

4 u - 1 = 0 : u = \frac{1}{4}

4 u - 1 = 0

Move

1

to the right side

4 u - 1 = 0

Add

1

to both sides

4 u - 1 + 1 = 0 + 1

Simplify

4 u = 1

4 u = 1

Divide both sides by

4

4 u = 1

Divide both sides by

4

\frac{4 u}{4} = \frac{1}{4}

Simplify

u = \frac{1}{4}

u = \frac{1}{4}

The following points are undefined

u = \frac{1}{4}

Combine undefined points with solutions:

u = \frac{1}{3}

Substitute back

u = sin (2 x)

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

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

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

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

Combine all the solutions

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

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for (sin(2x))/(4sin(2x)-1)=1 ?
The general solution for (sin(2x))/(4sin(2x)-1)=1 is x=(0.33983…)/2+pin,x= pi/2-(0.33983…)/2+pin