What is the general solution for -3cos(pi/2-θ)=tan(θ) ?

The general solution for -3cos(pi/2-θ)=tan(θ) is θ=2pin,θ=pi+2pin,θ=1.91063…+2pin,θ=-1.91063…+2pin

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Popular Trigonometry >

-3cos(pi/2-θ)=tan(θ)

Solution

- 3 cos (\frac{π}{2} - θ) = tan (θ)

Solution

θ = 2 πn, θ = π + 2 πn, θ = 1.91063 \dots + 2 πn, θ = - 1.91063 \dots + 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n, θ = 109.47122 \dots^{\circ} + 36 0^{\circ} n, θ = - 109.47122 \dots^{\circ} + 36 0^{\circ} n

Solution steps

- 3 cos (\frac{π}{2} - θ) = tan (θ)

Rewrite using trig identities

- 3 cos (\frac{π}{2} - θ) = tan (θ)

Rewrite using trig identities

cos (\frac{π}{2} - θ)

Use the Angle Difference identity:

cos (s - t) = cos (s) cos (t) + sin (s) sin (t)

= cos (\frac{π}{2}) cos (θ) + sin (\frac{π}{2}) sin (θ)

Simplify

cos (\frac{π}{2}) cos (θ) + sin (\frac{π}{2}) sin (θ) : sin (θ)

cos (\frac{π}{2}) cos (θ) + sin (\frac{π}{2}) sin (θ)

cos (\frac{π}{2}) cos (θ) = 0

cos (\frac{π}{2}) cos (θ)

Simplify

cos (\frac{π}{2}) : 0

cos (\frac{π}{2})

Use the following trivial identity:

cos (\frac{π}{2}) = 0

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}

= 0

= 0 \cdot cos (θ)

Apply rule

0 \cdot a = 0

= 0

sin (\frac{π}{2}) sin (θ) = sin (θ)

sin (\frac{π}{2}) sin (θ)

Simplify

sin (\frac{π}{2}) : 1

sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 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}

= 1

= 1 \cdot sin (θ)

Multiply:

1 \cdot sin (θ) = sin (θ)

= sin (θ)

= 0 + sin (θ)

0 + sin (θ) = sin (θ)

= sin (θ)

= sin (θ)

- 3 sin (θ) = tan (θ)

- 3 sin (θ) = tan (θ)

Subtract

tan (θ)

from both sides

- 3 sin (θ) - tan (θ) = 0

Express with sin, cos

- tan (θ) - 3 sin (θ)

Use the basic trigonometric identity:

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

= - \frac{sin ( θ )}{cos ( θ )} - 3 sin (θ)

Simplify

- \frac{sin ( θ )}{cos ( θ )} - 3 sin (θ) : \frac{- sin ( θ ) - 3 sin ( θ ) cos ( θ )}{cos ( θ )}

- \frac{sin ( θ )}{cos ( θ )} - 3 sin (θ)

Convert element to fraction:

3 sin (θ) = \frac{3 sin ( θ ) cos ( θ )}{cos ( θ )}

= - \frac{sin ( θ )}{cos ( θ )} - \frac{3 sin ( θ ) cos ( θ )}{cos ( θ )}

Since the denominators are equal, combine the fractions:

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

= \frac{- sin ( θ ) - 3 sin ( θ ) cos ( θ )}{cos ( θ )}

= \frac{- sin ( θ ) - 3 sin ( θ ) cos ( θ )}{cos ( θ )}

\frac{- sin ( θ ) - 3 cos ( θ ) sin ( θ )}{cos ( θ )} = 0

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

- sin (θ) - 3 cos (θ) sin (θ) = 0

Factor

- sin (θ) - 3 cos (θ) sin (θ) : - sin (θ) (3 cos (θ) + 1)

- sin (θ) - 3 cos (θ) sin (θ)

Factor out common term

- sin (θ)

= - sin (θ) (1 + 3 cos (θ))

- sin (θ) (3 cos (θ) + 1) = 0

Solving each part separately

sin (θ) = 0 or 3 cos (θ) + 1 = 0

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = 0

General solutions for

sin (θ) = 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}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

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

3 cos (θ) + 1 = 0

Move

1

to the right side

3 cos (θ) + 1 = 0

Subtract

1

from both sides

3 cos (θ) + 1 - 1 = 0 - 1

Simplify

3 cos (θ) = - 1

3 cos (θ) = - 1

Divide both sides by

3

3 cos (θ) = - 1

Divide both sides by

3

\frac{3 cos ( θ )}{3} = \frac{- 1}{3}

Simplify

cos (θ) = - \frac{1}{3}

cos (θ) = - \frac{1}{3}

Apply trig inverse properties

cos (θ) = - \frac{1}{3}

General solutions for

cos (θ) = - \frac{1}{3}

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

θ = arccos (- \frac{1}{3}) + 2 πn, θ = - arccos (- \frac{1}{3}) + 2 πn

θ = arccos (- \frac{1}{3}) + 2 πn, θ = - arccos (- \frac{1}{3}) + 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn, θ = arccos (- \frac{1}{3}) + 2 πn, θ = - arccos (- \frac{1}{3}) + 2 πn

Show solutions in decimal form

θ = 2 πn, θ = π + 2 πn, θ = 1.91063 \dots + 2 πn, θ = - 1.91063 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for -3cos(pi/2-θ)=tan(θ) ?
The general solution for -3cos(pi/2-θ)=tan(θ) is θ=2pin,θ=pi+2pin,θ=1.91063…+2pin,θ=-1.91063…+2pin