What is the general solution for 5sin^2(x)=9tan(x) ?

The general solution for 5sin^2(x)=9tan(x) is x=2pin,x=pi+2pin

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5sin^2(x)=9tan(x)

Solution

5 sin^{2} (x) = 9 tan (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

5 sin^{2} (x) = 9 tan (x)

Subtract

9 tan (x)

from both sides

5 sin^{2} (x) - 9 tan (x) = 0

Express with sin, cos

5 sin^{2} (x) - 9 tan (x)

Use the basic trigonometric identity:

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

= 5 sin^{2} (x) - 9 \cdot \frac{sin ( x )}{cos ( x )}

Simplify

5 sin^{2} (x) - 9 \cdot \frac{sin ( x )}{cos ( x )} : \frac{5 sin ^{2} ( x ) cos ( x ) - 9 sin ( x )}{cos ( x )}

5 sin^{2} (x) - 9 \cdot \frac{sin ( x )}{cos ( x )}

Multiply

9 \cdot \frac{sin ( x )}{cos ( x )} : \frac{9 sin ( x )}{cos ( x )}

9 \cdot \frac{sin ( x )}{cos ( x )}

Multiply fractions:

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

= \frac{sin ( x ) \cdot 9}{cos ( x )}

= 5 sin^{2} (x) - \frac{9 sin ( x )}{cos ( x )}

Convert element to fraction:

5 sin^{2} (x) = \frac{5 sin ^{2} ( x ) cos ( x )}{cos ( x )}

= \frac{5 sin ^{2} ( x ) cos ( x )}{cos ( x )} - \frac{sin ( x ) \cdot 9}{cos ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{5 sin ^{2} ( x ) cos ( x ) - sin ( x ) \cdot 9}{cos ( x )}

= \frac{5 sin ^{2} ( x ) cos ( x ) - 9 sin ( x )}{cos ( x )}

\frac{- 9 sin ( x ) + 5 cos ( x ) sin ^{2} ( x )}{cos ( x )} = 0

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

- 9 sin (x) + 5 cos (x) sin^{2} (x) = 0

Factor

- 9 sin (x) + 5 cos (x) sin^{2} (x) : sin (x) (- 9 + 5 sin (x) cos (x))

- 9 sin (x) + 5 cos (x) sin^{2} (x)

Apply exponent rule:

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

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

= - 9 sin (x) + 5 sin (x) sin (x)

Factor out common term

sin (x)

= sin (x) (- 9 + 5 sin (x) cos (x))

sin (x) (- 9 + 5 sin (x) cos (x)) = 0

Solving each part separately

sin (x) = 0 or - 9 + 5 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

- 9 + 5 sin (x) cos (x) = 0 :

No Solution

- 9 + 5 sin (x) cos (x) = 0

Rewrite using trig identities

- 9 + 5 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}

= - 9 + 5 \cdot \frac{sin ( 2 x )}{2}

- 9 + 5 \cdot \frac{sin ( 2 x )}{2} = 0

Move

9

to the right side

- 9 + 5 \cdot \frac{sin ( 2 x )}{2} = 0

Add

9

to both sides

- 9 + 5 \cdot \frac{sin ( 2 x )}{2} + 9 = 0 + 9

Simplify

5 \cdot \frac{sin ( 2 x )}{2} = 9

5 \cdot \frac{sin ( 2 x )}{2} = 9

Refine

5 \cdot \frac{sin ( 2 x )}{2} : \frac{5 sin ( 2 x )}{2}

5 \cdot \frac{sin ( 2 x )}{2}

Multiply fractions:

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

= \frac{sin ( 2 x ) \cdot 5}{2}

\frac{5 sin ( 2 x )}{2} = 9

Multiply both sides by

2

\frac{5 sin ( 2 x )}{2} = 9

Multiply both sides by

2

\frac{5 sin ( 2 x )}{2} \cdot 2 = 9 \cdot 2

Simplify

\frac{5 sin ( 2 x )}{2} \cdot 2 = 9 \cdot 2

Simplify

\frac{5 sin ( 2 x )}{2} \cdot 2 : 5 sin (2 x)

\frac{5 sin ( 2 x )}{2} \cdot 2

Multiply fractions:

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

= \frac{5 sin ( 2 x ) \cdot 2}{2}

Cancel the common factor:

2

= 5 sin (2 x)

Simplify

9 \cdot 2 : 18

9 \cdot 2

Multiply the numbers:

9 \cdot 2 = 18

= 18

5 sin (2 x) = 18

5 sin (2 x) = 18

5 sin (2 x) = 18

Divide both sides by

5

5 sin (2 x) = 18

Divide both sides by

5

\frac{5 sin ( 2 x )}{5} = \frac{18}{5}

Simplify

sin (2 x) = \frac{18}{5}

sin (2 x) = \frac{18}{5}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = 2 πn, x = π + 2 πn

Popular Examples

cos(θ)= 5/4

cos (θ) = \frac{5}{4}

cos(x)=sin(2x),0<= x<= 2pi

cos (x) = sin (2 x), 0 \leq x \leq 2 π

tan(θ)(sin(θ)-1)=0

tan (θ) (sin (θ) - 1) = 0

sin(θ)-4=-5

sin (θ) - 4 = - 5

sin(θ)-4=-3

sin (θ) - 4 = - 3

Frequently Asked Questions (FAQ)

What is the general solution for 5sin^2(x)=9tan(x) ?
The general solution for 5sin^2(x)=9tan(x) is x=2pin,x=pi+2pin