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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

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

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

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

Factor

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

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

Apply exponent rule:

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

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

= - sin (x) + sin (x) sin (x)

Factor out common term

sin (x)

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

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

Solving each part separately

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

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

No Solution

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

Rewrite using trig identities

- 1 + sin (x) cos (x)

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

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

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

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

Move

1

to the right side

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

Add

1

to both sides

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

Simplify

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

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

Multiply both sides by

2

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

Multiply both sides by

2

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

Simplify

sin (2 x) = 2

sin (2 x) = 2

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = 2 πn, x = π + 2 πn

Popular Examples

6sin^3(x)+7sin^2(x)-sin(x)-2=0

6 sin^{3} (x) + 7 sin^{2} (x) - sin (x) - 2 = 0

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

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

tan(x)= 1/(sec(x))

tan (x) = \frac{1}{sec ( x )}

2*cos(2x)-1=0

2 \cdot cos (2 x) - 1 = 0

sin^2(x)=((7m-2))/8

sin^{2} (x) = \frac{( 7 m - 2 )}{8}

Frequently Asked Questions (FAQ)

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