en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(tan^2(θ)-9)(2cos(θ)+1)=0

Solution

(tan^{2} (θ) - 9) (2 cos (θ) + 1) = 0

Solution

θ = 1.24904 \dots + πn, θ = - 1.24904 \dots + πn, θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

+1

Degrees

θ = 71.56505 \dots^{\circ} + 18 0^{\circ} n, θ = - 71.56505 \dots^{\circ} + 18 0^{\circ} n, θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

(tan^{2} (θ) - 9) (2 cos (θ) + 1) = 0

Solving each part separately

tan^{2} (θ) - 9 = 0 or 2 cos (θ) + 1 = 0

tan^{2} (θ) - 9 = 0 : θ = arctan (3) + πn, θ = arctan (- 3) + πn

tan^{2} (θ) - 9 = 0

Solve by substitution

tan^{2} (θ) - 9 = 0

Let:

tan (θ) = u

u^{2} - 9 = 0

u^{2} - 9 = 0 : u = 3, u = - 3

u^{2} - 9 = 0

Move

9

to the right side

u^{2} - 9 = 0

Add

9

to both sides

u^{2} - 9 + 9 = 0 + 9

Simplify

u^{2} = 9

u^{2} = 9

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 9, u = - 9

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

- 9 = - 3

- 9

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

= - 3

u = 3, u = - 3

Substitute back

u = tan (θ)

tan (θ) = 3, tan (θ) = - 3

tan (θ) = 3, tan (θ) = - 3

tan (θ) = 3 : θ = arctan (3) + πn

tan (θ) = 3

Apply trig inverse properties

tan (θ) = 3

General solutions for

tan (θ) = 3

tan (x) = a \Rightarrow x = arctan (a) + πn

θ = arctan (3) + πn

θ = arctan (3) + πn

tan (θ) = - 3 : θ = arctan (- 3) + πn

tan (θ) = - 3

Apply trig inverse properties

tan (θ) = - 3

General solutions for

tan (θ) = - 3

tan (x) = - a \Rightarrow x = arctan (- a) + πn

θ = arctan (- 3) + πn

θ = arctan (- 3) + πn

Combine all the solutions

θ = arctan (3) + πn, θ = arctan (- 3) + πn

2 cos (θ) + 1 = 0 : θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

2 cos (θ) + 1 = 0

Move

1

to the right side

2 cos (θ) + 1 = 0

Subtract

1

from both sides

2 cos (θ) + 1 - 1 = 0 - 1

Simplify

2 cos (θ) = - 1

2 cos (θ) = - 1

Divide both sides by

2

2 cos (θ) = - 1

Divide both sides by

2

\frac{2 cos ( θ )}{2} = \frac{- 1}{2}

Simplify

cos (θ) = - \frac{1}{2}

cos (θ) = - \frac{1}{2}

General solutions for

cos (θ) = - \frac{1}{2}

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

Combine all the solutions

θ = arctan (3) + πn, θ = arctan (- 3) + πn, θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

Show solutions in decimal form

θ = 1.24904 \dots + πn, θ = - 1.24904 \dots + πn, θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

Popular Examples

2sin(x)+cos(x)=1

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

sin(θ)cos(θ)=-1/2

sin (θ) cos (θ) = - \frac{1}{2}

2cos^2(x)-3cos(2x)=0

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

16 cos^{2} (x) = 4

3 sin (2 x - 1) - 1 = 0