What is the general solution for 3cos(x)tan^2(x)=cos(x) ?

The general solution for 3cos(x)tan^2(x)=cos(x) is x=(5pi)/6+pin,x= pi/6+pin

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3cos(x)tan^2(x)=cos(x)

Solution

3 cos (x) tan^{2} (x) = cos (x)

Solution

x = \frac{5 π}{6} + πn, x = \frac{π}{6} + πn

Degrees

x = 15 0^{\circ} + 18 0^{\circ} n, x = 3 0^{\circ} + 18 0^{\circ} n

Solution steps

3 cos (x) tan^{2} (x) = cos (x)

Subtract

cos (x)

from both sides

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

Factor

3 tan^{2} (x) cos (x) - cos (x) : cos (x) (3 tan (x) + 1) (3 tan (x) - 1)

3 tan^{2} (x) cos (x) - cos (x)

Factor out common term

cos (x)

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

Factor

3 tan^{2} (x) - 1 : (3 tan (x) + 1) (3 tan (x) - 1)

3 tan^{2} (x) - 1

Rewrite

3 tan^{2} (x) - 1

(3 tan (x))^{2} - 1^{2}

3 tan^{2} (x) - 1

Apply radical rule:

a = (a)^{2}

3 = (3)^{2}

= (3)^{2} tan^{2} (x) - 1

Rewrite

1

1^{2}

= (3)^{2} tan^{2} (x) - 1^{2}

Apply exponent rule:

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

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

= (3 tan (x))^{2} - 1^{2}

= (3 tan (x))^{2} - 1^{2}

Apply Difference of Two Squares Formula:

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

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

= (3 tan (x) + 1) (3 tan (x) - 1)

= cos (x) (3 tan (x) + 1) (3 tan (x) - 1)

cos (x) (3 tan (x) + 1) (3 tan (x) - 1) = 0

Solving each part separately

cos (x) = 0 or 3 tan (x) + 1 = 0 or 3 tan (x) - 1 = 0

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

cos (x) = 0

General solutions for

cos (x) = 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}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

3 tan (x) + 1 = 0 : x = \frac{5 π}{6} + πn

3 tan (x) + 1 = 0

Move

1

to the right side

3 tan (x) + 1 = 0

Subtract

1

from both sides

3 tan (x) + 1 - 1 = 0 - 1

Simplify

3 tan (x) = - 1

3 tan (x) = - 1

Divide both sides by

3

3 tan (x) = - 1

Divide both sides by

3

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

Simplify

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

Simplify

\frac{3 tan ( x )}{3} : tan (x)

\frac{3 tan ( x )}{3}

Cancel the common factor:

3

= tan (x)

Simplify

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

\frac{- 1}{3}

Apply the fraction rule:

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

= - \frac{1}{3}

Rationalize

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

- \frac{1}{3}

Multiply by the conjugate

\frac{3}{3}

= - \frac{1 \cdot 3}{3 3}

1 \cdot 3 = 3

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= - \frac{3}{3}

= - \frac{3}{3}

tan (x) = - \frac{3}{3}

tan (x) = - \frac{3}{3}

tan (x) = - \frac{3}{3}

General solutions for

tan (x) = - \frac{3}{3}

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{5 π}{6} + πn

x = \frac{5 π}{6} + πn

3 tan (x) - 1 = 0 : x = \frac{π}{6} + πn

3 tan (x) - 1 = 0

Move

1

to the right side

3 tan (x) - 1 = 0

Add

1

to both sides

3 tan (x) - 1 + 1 = 0 + 1

Simplify

3 tan (x) = 1

3 tan (x) = 1

Divide both sides by

3

3 tan (x) = 1

Divide both sides by

3

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

Simplify

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

Simplify

\frac{3 tan ( x )}{3} : tan (x)

\frac{3 tan ( x )}{3}

Cancel the common factor:

3

= tan (x)

Simplify

\frac{1}{3} : \frac{3}{3}

\frac{1}{3}

Multiply by the conjugate

\frac{3}{3}

= \frac{1 \cdot 3}{3 3}

1 \cdot 3 = 3

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= \frac{3}{3}

tan (x) = \frac{3}{3}

tan (x) = \frac{3}{3}

tan (x) = \frac{3}{3}

General solutions for

tan (x) = \frac{3}{3}

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{6} + πn

x = \frac{π}{6} + πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{5 π}{6} + πn, x = \frac{π}{6} + πn

Since the equation is undefined for:

\frac{π}{2} + 2 πn, \frac{3 π}{2} + 2 πn

x = \frac{5 π}{6} + πn, x = \frac{π}{6} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 3cos(x)tan^2(x)=cos(x) ?
The general solution for 3cos(x)tan^2(x)=cos(x) is x=(5pi)/6+pin,x= pi/6+pin