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

The general solution for tan(2x)sin(x)-sin(x)-tan(2x)+1=0 is x= pi/8+(pin)/2 ,x= pi/2+2pin

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

Solution

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

Solution

x = \frac{π}{8} + \frac{πn}{2}, x = \frac{π}{2} + 2 πn

Degrees

x = 22. 5^{\circ} + 9 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Factor

tan (2 x) sin (x) - sin (x) - tan (2 x) + 1 : (tan (2 x) - 1) (sin (x) - 1)

tan (2 x) sin (x) - sin (x) - tan (2 x) + 1

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

Factor out

- 1

from

- tan (2 x) + 1 : - (tan (2 x) - 1)

- tan (2 x) + 1

Factor out common term

- 1

= - (tan (2 x) - 1)

Factor out

sin (x)

from

sin (x) tan (2 x) - sin (x) : sin (x) (tan (2 x) - 1)

sin (x) tan (2 x) - sin (x)

Factor out common term

sin (x)

= sin (x) (tan (2 x) - 1)

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

Factor out common term

tan (2 x) - 1

= (tan (2 x) - 1) (sin (x) - 1)

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

Solving each part separately

tan (2 x) - 1 = 0 or sin (x) - 1 = 0

tan (2 x) - 1 = 0 : x = \frac{π}{8} + \frac{πn}{2}

tan (2 x) - 1 = 0

Move

1

to the right side

tan (2 x) - 1 = 0

Add

1

to both sides

tan (2 x) - 1 + 1 = 0 + 1

Simplify

tan (2 x) = 1

tan (2 x) = 1

General solutions for

tan (2 x) = 1

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

2 x = \frac{π}{4} + πn

2 x = \frac{π}{4} + πn

Solve

2 x = \frac{π}{4} + πn : x = \frac{π}{8} + \frac{πn}{2}

2 x = \frac{π}{4} + πn

Divide both sides by

2

2 x = \frac{π}{4} + πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{π}{4}}{2} + \frac{πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{π}{4}}{2} + \frac{πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{π}{4}}{2} + \frac{πn}{2} : \frac{π}{8} + \frac{πn}{2}

\frac{\frac{π}{4}}{2} + \frac{πn}{2}

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{π}{8}

= \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

sin (x) - 1 = 0 : x = \frac{π}{2} + 2 πn

sin (x) - 1 = 0

Move

1

to the right side

sin (x) - 1 = 0

Add

1

to both sides

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

Simplify

sin (x) = 1

sin (x) = 1

General solutions for

sin (x) = 1

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{8} + \frac{πn}{2}, x = \frac{π}{2} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for tan(2x)sin(x)-sin(x)-tan(2x)+1=0 ?
The general solution for tan(2x)sin(x)-sin(x)-tan(2x)+1=0 is x= pi/8+(pin)/2 ,x= pi/2+2pin