What is the general solution for 3sin(x)tan(x)=8 ?

The general solution for 3sin(x)tan(x)=8 is x=1.23095…+2pin,x=2pi-1.23095…+2pin

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3sin(x)tan(x)=8

Solution

3 sin (x) tan (x) = 8

Solution

x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn

Degrees

x = 70.52877 \dots^{\circ} + 36 0^{\circ} n, x = 289.47122 \dots^{\circ} + 36 0^{\circ} n

Solution steps

3 sin (x) tan (x) = 8

Subtract

8

from both sides

3 sin (x) tan (x) - 8 = 0

Express with sin, cos

3 sin (x) \frac{sin ( x )}{cos ( x )} - 8 = 0

Simplify

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

3 sin (x) \frac{sin ( x )}{cos ( x )} - 8

3 sin (x) \frac{sin ( x )}{cos ( x )} = \frac{3 sin ^{2} ( x )}{cos ( x )}

3 sin (x) \frac{sin ( x )}{cos ( x )}

Multiply fractions:

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

= \frac{sin ( x ) \cdot 3 sin ( x )}{cos ( x )}

sin (x) \cdot 3 sin (x) = 3 sin^{2} (x)

sin (x) \cdot 3 sin (x)

Apply exponent rule:

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

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

= 3 sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= 3 sin^{2} (x)

= \frac{3 sin ^{2} ( x )}{cos ( x )}

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

Convert element to fraction:

8 = \frac{8 cos ( x )}{cos ( x )}

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

Since the denominators are equal, combine the fractions:

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

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

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

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

3 sin^{2} (x) - 8 cos (x) = 0

Add

8 cos (x)

to both sides

3 sin^{2} (x) = 8 cos (x)

Square both sides

(3 sin^{2} (x))^{2} = (8 cos (x))^{2}

Subtract

(8 cos (x))^{2}

from both sides

9 sin^{4} (x) - 64 cos^{2} (x) = 0

Factor

9 sin^{4} (x) - 64 cos^{2} (x) : (3 sin^{2} (x) + 8 cos (x)) (3 sin^{2} (x) - 8 cos (x))

9 sin^{4} (x) - 64 cos^{2} (x)

Rewrite

9 sin^{4} (x) - 64 cos^{2} (x)

(3 sin^{2} (x))^{2} - (8 cos (x))^{2}

9 sin^{4} (x) - 64 cos^{2} (x)

Rewrite

9

3^{2}

= 3^{2} sin^{4} (x) - 64 cos^{2} (x)

Rewrite

64

8^{2}

= 3^{2} sin^{4} (x) - 8^{2} cos^{2} (x)

Apply exponent rule:

a^{b c} = (a^{b})^{c}

sin^{4} (x) = (sin^{2} (x))^{2}

= 3^{2} (sin^{2} (x))^{2} - 8^{2} cos^{2} (x)

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

3^{2} (sin^{2} (x))^{2} = (3 sin^{2} (x))^{2}

= (3 sin^{2} (x))^{2} - 8^{2} cos^{2} (x)

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

8^{2} cos^{2} (x) = (8 cos (x))^{2}

= (3 sin^{2} (x))^{2} - (8 cos (x))^{2}

= (3 sin^{2} (x))^{2} - (8 cos (x))^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

(3 sin^{2} (x))^{2} - (8 cos (x))^{2} = (3 sin^{2} (x) + 8 cos (x)) (3 sin^{2} (x) - 8 cos (x))

= (3 sin^{2} (x) + 8 cos (x)) (3 sin^{2} (x) - 8 cos (x))

(3 sin^{2} (x) + 8 cos (x)) (3 sin^{2} (x) - 8 cos (x)) = 0

Solving each part separately

3 sin^{2} (x) + 8 cos (x) = 0 or 3 sin^{2} (x) - 8 cos (x) = 0

3 sin^{2} (x) + 8 cos (x) = 0 : x = arccos (- \frac{1}{3}) + 2 πn, x = - arccos (- \frac{1}{3}) + 2 πn

3 sin^{2} (x) + 8 cos (x) = 0

Rewrite using trig identities

3 sin^{2} (x) + 8 cos (x)

Use the Pythagorean identity:

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

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

= 3 (1 - cos^{2} (x)) + 8 cos (x)

(1 - cos^{2} (x)) \cdot 3 + 8 cos (x) = 0

Solve by substitution

(1 - cos^{2} (x)) \cdot 3 + 8 cos (x) = 0

Let:

cos (x) = u

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

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

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 )} : - \frac{1}{3}

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

Remove parentheses:

(- a) = - a

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

Add/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}

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

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

Remove parentheses:

(- a) = - a

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

Subtract 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

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

cos (x) = - \frac{1}{3}, cos (x) = 3

cos (x) = - \frac{1}{3}, cos (x) = 3

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

cos (x) = - \frac{1}{3}

Apply trig inverse properties

cos (x) = - \frac{1}{3}

General solutions for

cos (x) = - \frac{1}{3}

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

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

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

cos (x) = 3 :

No Solution

cos (x) = 3

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

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

3 sin^{2} (x) - 8 cos (x) = 0 : x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

3 sin^{2} (x) - 8 cos (x) = 0

Rewrite using trig identities

3 sin^{2} (x) - 8 cos (x)

Use the Pythagorean identity:

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

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

= 3 (1 - cos^{2} (x)) - 8 cos (x)

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

Solve by substitution

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

Let:

cos (x) = 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 = cos (x)

cos (x) = - 3, cos (x) = \frac{1}{3}

cos (x) = - 3, cos (x) = \frac{1}{3}

cos (x) = - 3 :

No Solution

cos (x) = - 3

- 1 \leq cos (x) \leq 1

No Solution

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

cos (x) = \frac{1}{3}

Apply trig inverse properties

cos (x) = \frac{1}{3}

General solutions for

cos (x) = \frac{1}{3}

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

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

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

Combine all the solutions

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

Combine all the solutions

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

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

3 sin (x) tan (x) = 8

Remove the ones that don't agree with the equation.

Check the solution

arccos (- \frac{1}{3}) + 2 πn :

False

arccos (- \frac{1}{3}) + 2 πn

Plug in

n = 1

arccos (- \frac{1}{3}) + 2 π 1

For

3 sin (x) tan (x) = 8

plug in

x = arccos (- \frac{1}{3}) + 2 π 1

3 sin (arccos (- \frac{1}{3}) + 2 π 1) tan (arccos (- \frac{1}{3}) + 2 π 1) = 8

Refine

- 8 = 8

\Rightarrow False

Check the solution

- arccos (- \frac{1}{3}) + 2 πn :

False

- arccos (- \frac{1}{3}) + 2 πn

Plug in

n = 1

- arccos (- \frac{1}{3}) + 2 π 1

For

3 sin (x) tan (x) = 8

plug in

x = - arccos (- \frac{1}{3}) + 2 π 1

3 sin (- arccos (- \frac{1}{3}) + 2 π 1) tan (- arccos (- \frac{1}{3}) + 2 π 1) = 8

Refine

- 8 = 8

\Rightarrow False

Check the solution

arccos (\frac{1}{3}) + 2 πn :

True

arccos (\frac{1}{3}) + 2 πn

Plug in

n = 1

arccos (\frac{1}{3}) + 2 π 1

For

3 sin (x) tan (x) = 8

plug in

x = arccos (\frac{1}{3}) + 2 π 1

3 sin (arccos (\frac{1}{3}) + 2 π 1) tan (arccos (\frac{1}{3}) + 2 π 1) = 8

Refine

8 = 8

\Rightarrow True

Check the solution

2 π - arccos (\frac{1}{3}) + 2 πn :

True

2 π - arccos (\frac{1}{3}) + 2 πn

Plug in

n = 1

2 π - arccos (\frac{1}{3}) + 2 π 1

For

3 sin (x) tan (x) = 8

plug in

x = 2 π - arccos (\frac{1}{3}) + 2 π 1

3 sin (2 π - arccos (\frac{1}{3}) + 2 π 1) tan (2 π - arccos (\frac{1}{3}) + 2 π 1) = 8

Refine

8 = 8

\Rightarrow True

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

Show solutions in decimal form

x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 3sin(x)tan(x)=8 ?
The general solution for 3sin(x)tan(x)=8 is x=1.23095…+2pin,x=2pi-1.23095…+2pin