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Popular >

tan^2(x)sin(x)=sin(x),0<= x<= 2pi

Solution

tan^{2} (x) sin (x) = sin (x), 0 \leq x \leq 2 π

Solution

x = 0, x = π, x = 2 π, x = \frac{3 π}{4}, x = \frac{7 π}{4}, x = \frac{π}{4}, x = \frac{5 π}{4}

+1

Degrees

x = 0^{\circ}, x = 18 0^{\circ}, x = 36 0^{\circ}, x = 13 5^{\circ}, x = 31 5^{\circ}, x = 4 5^{\circ}, x = 22 5^{\circ}

Solution steps

tan^{2} (x) sin (x) = sin (x), 0 \leq x \leq 2 π

Subtract

sin (x)

from both sides

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

as

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, 0 \leq x \leq 2 π : x = 0, x = π, x = 2 π

sin (x) = 0, 0 \leq x \leq 2 π

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

Solutions for the range

0 \leq x \leq 2 π

x = 0, x = π, x = 2 π

tan (x) + 1 = 0, 0 \leq x \leq 2 π : x = \frac{3 π}{4}, x = \frac{7 π}{4}

tan (x) + 1 = 0, 0 \leq x \leq 2 π

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

Solutions for the range

0 \leq x \leq 2 π

x = \frac{3 π}{4}, x = \frac{7 π}{4}

tan (x) - 1 = 0, 0 \leq x \leq 2 π : x = \frac{π}{4}, x = \frac{5 π}{4}

tan (x) - 1 = 0, 0 \leq x \leq 2 π

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

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{4}, x = \frac{5 π}{4}

Combine all the solutions

x = 0, x = π, x = 2 π, x = \frac{3 π}{4}, x = \frac{7 π}{4}, x = \frac{π}{4}, x = \frac{5 π}{4}

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