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

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

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Popular Trigonometry >

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

Solution

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

Solution

x = 2 πn, x = π + 2 πn

Degrees

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

Solution steps

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

Subtract

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

from both sides

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

Express with sin, cos

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

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= - sin (x) + (\frac{sin ( x )}{cos ( x )})^{2} - sin^{2} (x) (\frac{sin ( x )}{cos ( x )})^{2}

Simplify

- sin (x) + (\frac{sin ( x )}{cos ( x )})^{2} - sin^{2} (x) (\frac{sin ( x )}{cos ( x )})^{2} : \frac{- cos ^{2} ( x ) sin ( x ) + sin ^{2} ( x ) - sin ^{4} ( x )}{cos ^{2} ( x )}

- sin (x) + (\frac{sin ( x )}{cos ( x )})^{2} - sin^{2} (x) (\frac{sin ( x )}{cos ( x )})^{2}

(\frac{sin ( x )}{cos ( x )})^{2} = \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

(\frac{sin ( x )}{cos ( x )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

sin^{2} (x) (\frac{sin ( x )}{cos ( x )})^{2} = \frac{sin ^{4} ( x )}{cos ^{2} ( x )}

sin^{2} (x) (\frac{sin ( x )}{cos ( x )})^{2}

(\frac{sin ( x )}{cos ( x )})^{2} = \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

(\frac{sin ( x )}{cos ( x )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

= \frac{sin ^{2} ( x )}{cos ^{2} ( x )} sin^{2} (x)

Multiply fractions:

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

= \frac{sin ^{2} ( x ) sin ^{2} ( x )}{cos ^{2} ( x )}

sin^{2} (x) sin^{2} (x) = sin^{4} (x)

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

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

sin^{2} (x) sin^{2} (x) = sin^{2 + 2} (x)

= sin^{2 + 2} (x)

Add the numbers:

2 + 2 = 4

= sin^{4} (x)

= \frac{sin ^{4} ( x )}{cos ^{2} ( x )}

= - sin (x) + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} - \frac{sin ^{4} ( x )}{cos ^{2} ( x )}

Combine the fractions

\frac{sin ^{2} ( x )}{cos ^{2} ( x )} - \frac{sin ^{4} ( x )}{cos ^{2} ( x )} : \frac{sin ^{2} ( x ) - sin ^{4} ( x )}{cos ^{2} ( x )}

Apply rule

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

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

= - sin (x) + \frac{sin ^{2} ( x ) - sin ^{4} ( x )}{cos ^{2} ( x )}

Convert element to fraction:

sin (x) = \frac{sin ( x ) cos ^{2} ( x )}{cos ^{2} ( x )}

= - \frac{sin ( x ) cos ^{2} ( x )}{cos ^{2} ( x )} + \frac{sin ^{2} ( x ) - sin ^{4} ( x )}{cos ^{2} ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{- sin ( x ) cos ^{2} ( x ) + sin ^{2} ( x ) - sin ^{4} ( x )}{cos ^{2} ( x )}

= \frac{- cos ^{2} ( x ) sin ( x ) + sin ^{2} ( x ) - sin ^{4} ( x )}{cos ^{2} ( x )}

\frac{sin ^{2} ( x ) - sin ^{4} ( x ) - cos ^{2} ( x ) sin ( x )}{cos ^{2} ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

sin^{2} (x) - sin^{4} (x) - cos^{2} (x) sin (x) = 0

Factor

sin^{2} (x) - sin^{4} (x) - cos^{2} (x) sin (x) : sin (x) (sin (x) - sin^{3} (x) - cos^{2} (x))

sin^{2} (x) - sin^{4} (x) - cos^{2} (x) sin (x)

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

sin^{4} (x) = sin (x) sin^{3} (x), sin^{2} (x) = sin (x) sin (x)

= sin (x) sin (x) - sin (x) sin^{3} (x) - sin (x) cos^{2} (x)

Factor out common term

sin (x)

= sin (x) (sin (x) - sin^{3} (x) - cos^{2} (x))

sin (x) (sin (x) - sin^{3} (x) - cos^{2} (x)) = 0

Solving each part separately

sin (x) = 0 or sin (x) - sin^{3} (x) - cos^{2} (x) = 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

sin (x) - sin^{3} (x) - cos^{2} (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

sin (x) - sin^{3} (x) - cos^{2} (x) = 0

Rewrite using trig identities

- cos^{2} (x) + sin (x) - sin^{3} (x)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

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

= - (1 - sin^{2} (x)) + sin (x) - sin^{3} (x)

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

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

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

= - 1 + sin^{2} (x) + sin (x) - sin^{3} (x)

- 1 + sin (x) + sin^{2} (x) - sin^{3} (x) = 0

Solve by substitution

- 1 + sin (x) + sin^{2} (x) - sin^{3} (x) = 0

Let:

sin (x) = u

- 1 + u + u^{2} - u^{3} = 0

- 1 + u + u^{2} - u^{3} = 0 : u = 1, u = - 1

- 1 + u + u^{2} - u^{3} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

- u^{3} + u^{2} + u - 1 = 0

Factor

- u^{3} + u^{2} + u - 1 : - (u - 1)^{2} (u + 1)

- u^{3} + u^{2} + u - 1

Factor out common term

- 1

= - (u^{3} - u^{2} - u + 1)

Factor

u^{3} - u^{2} - u + 1 : (u - 1) (u + 1) (u - 1)

u^{3} - u^{2} - u + 1

= (u^{3} - u^{2}) + (- u + 1)

Factor out

- 1

from

- u + 1 : - (u - 1)

- u + 1

Factor out common term

- 1

= - (u - 1)

Factor out

u^{2}

from

u^{3} - u^{2} : u^{2} (u - 1)

u^{3} - u^{2}

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

u^{3} = u u^{2}

= u u^{2} - u^{2}

Factor out common term

u^{2}

= u^{2} (u - 1)

= - (u - 1) + u^{2} (u - 1)

Factor out common term

u - 1

= (u - 1) (u^{2} - 1)

Factor

u^{2} - 1 : (u + 1) (u - 1)

u^{2} - 1

Rewrite

1

1^{2}

= u^{2} - 1^{2}

Apply Difference of Two Squares Formula:

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

u^{2} - 1^{2} = (u + 1) (u - 1)

= (u + 1) (u - 1)

= (u - 1) (u + 1) (u - 1)

= - (u - 1) (u + 1) (u - 1)

Refine

= - (u - 1)^{2} (u + 1)

- (u - 1)^{2} (u + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u - 1 = 0 or u + 1 = 0

Solve

u - 1 = 0 : u = 1

u - 1 = 0

Move

1

to the right side

u - 1 = 0

Add

1

to both sides

u - 1 + 1 = 0 + 1

Simplify

u = 1

u = 1

Solve

u + 1 = 0 : u = - 1

u + 1 = 0

Move

1

to the right side

u + 1 = 0

Subtract

1

from both sides

u + 1 - 1 = 0 - 1

Simplify

u = - 1

u = - 1

The solutions are

u = 1, u = - 1

Substitute back

u = sin (x)

sin (x) = 1, sin (x) = - 1

sin (x) = 1, sin (x) = - 1

sin (x) = 1 : x = \frac{π}{2} + 2 πn

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

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

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{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

Combine all the solutions

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

Combine all the solutions

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

Since the equation is undefined for:

\frac{π}{2} + 2 πn, \frac{3 π}{2} + 2 πn

x = 2 πn, x = π + 2 πn

Graph

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Frequently Asked Questions (FAQ)

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