What is the general solution for 3cos(θ)=8tan(θ) ?

The general solution for 3cos(θ)=8tan(θ) is θ=0.33983…+2pin,θ=pi-0.33983…+2pin

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Popular >

3cos(θ)=8tan(θ)

Solution

3 cos (θ) = 8 tan (θ)

Solution

θ = 0.33983 \dots + 2 πn, θ = π - 0.33983 \dots + 2 πn

Degrees

θ = 19.47122 \dots^{\circ} + 36 0^{\circ} n, θ = 160.52877 \dots^{\circ} + 36 0^{\circ} n

Solution steps

3 cos (θ) = 8 tan (θ)

Subtract

8 tan (θ)

from both sides

3 cos (θ) - 8 tan (θ) = 0

Express with sin, cos

3 cos (θ) - 8 tan (θ)

Use the basic trigonometric identity:

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

= 3 cos (θ) - 8 \cdot \frac{sin ( θ )}{cos ( θ )}

Simplify

3 cos (θ) - 8 \cdot \frac{sin ( θ )}{cos ( θ )} : \frac{3 cos ^{2} ( θ ) - 8 sin ( θ )}{cos ( θ )}

3 cos (θ) - 8 \cdot \frac{sin ( θ )}{cos ( θ )}

Multiply

8 \cdot \frac{sin ( θ )}{cos ( θ )} : \frac{8 sin ( θ )}{cos ( θ )}

8 \cdot \frac{sin ( θ )}{cos ( θ )}

Multiply fractions:

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

= \frac{sin ( θ ) \cdot 8}{cos ( θ )}

= 3 cos (θ) - \frac{8 sin ( θ )}{cos ( θ )}

Convert element to fraction:

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

= \frac{3 cos ( θ ) cos ( θ )}{cos ( θ )} - \frac{sin ( θ ) \cdot 8}{cos ( θ )}

Since the denominators are equal, combine the fractions:

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

= \frac{3 cos ( θ ) cos ( θ ) - sin ( θ ) \cdot 8}{cos ( θ )}

3 cos (θ) cos (θ) - sin (θ) \cdot 8 = 3 cos^{2} (θ) - 8 sin (θ)

3 cos (θ) cos (θ) - sin (θ) \cdot 8

3 cos (θ) cos (θ) = 3 cos^{2} (θ)

3 cos (θ) cos (θ)

Apply exponent rule:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= 3 cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= 3 cos^{2} (θ)

= 3 cos^{2} (θ) - 8 sin (θ)

= \frac{3 cos ^{2} ( θ ) - 8 sin ( θ )}{cos ( θ )}

= \frac{3 cos ^{2} ( θ ) - 8 sin ( θ )}{cos ( θ )}

\frac{3 cos ^{2} ( θ ) - 8 sin ( θ )}{cos ( θ )} = 0

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

3 cos^{2} (θ) - 8 sin (θ) = 0

Rewrite using trig identities

3 cos^{2} (θ) - 8 sin (θ)

Use the Pythagorean identity:

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

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

= 3 (1 - sin^{2} (θ)) - 8 sin (θ)

(1 - sin^{2} (θ)) \cdot 3 - 8 sin (θ) = 0

Solve by substitution

(1 - sin^{2} (θ)) \cdot 3 - 8 sin (θ) = 0

Let:

sin (θ) = u

(1 - u^{2}) \cdot 3 - 8 u = 0

(1 - u^{2}) \cdot 3 - 8 u = 0 : u = - 3, u = \frac{1}{3}

(1 - u^{2}) \cdot 3 - 8 u = 0

Expand

(1 - u^{2}) \cdot 3 - 8 u : 3 - 3 u^{2} - 8 u

(1 - u^{2}) \cdot 3 - 8 u

= 3 (1 - u^{2}) - 8 u

Expand

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

3 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = 3, b = 1, c = u^{2}

= 3 \cdot 1 - 3 u^{2}

Multiply the numbers:

3 \cdot 1 = 3

= 3 - 3 u^{2}

= 3 - 3 u^{2} - 8 u

3 - 3 u^{2} - 8 u = 0

Write in the standard form

a x^{2} + b x + c = 0

- 3 u^{2} - 8 u + 3 = 0

Solve with the quadratic formula

- 3 u^{2} - 8 u + 3 = 0

Quadratic Equation Formula:

For

a = - 3, b = - 8, c = 3

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 3 ) \cdot 3}{2 ( - 3 )}

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 3 ) \cdot 3}{2 ( - 3 )}

(- 8)^{2} - 4 (- 3) \cdot 3 = 10

(- 8)^{2} - 4 (- 3) \cdot 3

Apply rule

- (- a) = a

= (- 8)^{2} + 4 \cdot 3 \cdot 3

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 8)^{2} = 8^{2}

= 8^{2} + 4 \cdot 3 \cdot 3

Multiply the numbers:

4 \cdot 3 \cdot 3 = 36

= 8^{2} + 36

8^{2} = 64

= 64 + 36

Add the numbers:

64 + 36 = 100

= 100

Factor the number:

100 = 1 0^{2}

= 1 0^{2}

Apply radical rule:

1 0^{2} = 10

= 10

u_{1, 2} = \frac{- ( - 8 ) \pm 10}{2 ( - 3 )}

Separate the solutions

u_{1} = \frac{- ( - 8 ) + 10}{2 ( - 3 )}, u_{2} = \frac{- ( - 8 ) - 10}{2 ( - 3 )}

u = \frac{- ( - 8 ) + 10}{2 ( - 3 )} : - 3

\frac{- ( - 8 ) + 10}{2 ( - 3 )}

Remove parentheses:

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

= \frac{8 + 10}{- 2 \cdot 3}

Add the numbers:

8 + 10 = 18

= \frac{18}{- 2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{18}{- 6}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{18}{6}

Divide the numbers:

\frac{18}{6} = 3

= - 3

u = \frac{- ( - 8 ) - 10}{2 ( - 3 )} : \frac{1}{3}

\frac{- ( - 8 ) - 10}{2 ( - 3 )}

Remove parentheses:

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

= \frac{8 - 10}{- 2 \cdot 3}

Subtract the numbers:

8 - 10 = - 2

= \frac{- 2}{- 2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 2}{- 6}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{2}{6}

Cancel the common factor:

2

= \frac{1}{3}

The solutions to the quadratic equation are:

u = - 3, u = \frac{1}{3}

Substitute back

u = sin (θ)

sin (θ) = - 3, sin (θ) = \frac{1}{3}

sin (θ) = - 3, sin (θ) = \frac{1}{3}

sin (θ) = - 3 :

No Solution

sin (θ) = - 3

- 1 \leq sin (x) \leq 1

No Solution

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

sin (θ) = \frac{1}{3}

Apply trig inverse properties

sin (θ) = \frac{1}{3}

General solutions for

sin (θ) = \frac{1}{3}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

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

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

Combine all the solutions

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

Show solutions in decimal form

θ = 0.33983 \dots + 2 πn, θ = π - 0.33983 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 3cos(θ)=8tan(θ) ?
The general solution for 3cos(θ)=8tan(θ) is θ=0.33983…+2pin,θ=pi-0.33983…+2pin