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

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

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Popular >

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

Solution

tan^{2} (x) sin (x) - sin (x) = 0

Solution

x = 2 πn, x = π + 2 πn, x = \frac{3 π}{4} + πn, x = \frac{π}{4} + πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (x) sin (x) - sin (x) = 0

Factor

tan^{2} (x) sin (x) - sin (x) : sin (x) (tan (x) + 1) (tan (x) - 1)

tan^{2} (x) sin (x) - sin (x)

Factor out common term

sin (x)

= sin (x) (tan^{2} (x) - 1)

Factor

tan^{2} (x) - 1 : (tan (x) + 1) (tan (x) - 1)

tan^{2} (x) - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

tan^{2} (x) - 1^{2} = (tan (x) + 1) (tan (x) - 1)

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

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

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

Solving each part separately

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

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

tan (x) + 1 = 0 : x = \frac{3 π}{4} + πn

tan (x) + 1 = 0

Move

1

to the right side

tan (x) + 1 = 0

Subtract

1

from both sides

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

Simplify

tan (x) = - 1

tan (x) = - 1

General solutions for

tan (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}

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

tan (x) - 1 = 0 : x = \frac{π}{4} + πn

tan (x) - 1 = 0

Move

1

to the right side

tan (x) - 1 = 0

Add

1

to both sides

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

Simplify

tan (x) = 1

tan (x) = 1

General solutions for

tan (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}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = \frac{3 π}{4} + πn, x = \frac{π}{4} + πn

Popular Examples

2cos^2(t)-3cos(t)+1=0

2 cos^{2} (t) - 3 cos (t) + 1 = 0

sin(x+30)=cos(x)

sin (x + 3 0^{\circ}) = cos (x)

sin(pi/6 x)=0

sin (\frac{π}{6} x) = 0

sec(x)-2cos(x)=1

sec (x) - 2 cos (x) = 1

2sin(x)+3=0

2 sin (x) + 3 = 0

Frequently Asked Questions (FAQ)

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