What is the general solution for cos(x)tan^2(x)-3cos(x)=0 ?

The general solution for cos(x)tan^2(x)-3cos(x)=0 is x=(2pi)/3+pin,x= pi/3+pin

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

Solution

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

Solution

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

Degrees

x = 12 0^{\circ} + 18 0^{\circ} n, x = 6 0^{\circ} + 18 0^{\circ} n

Solution steps

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

Factor

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

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

Factor out common term

cos (x)

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

Factor

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

tan^{2} (x) - 3

Apply radical rule:

a = (a)^{2}

3 = (3)^{2}

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

Apply Difference of Two Squares Formula:

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

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

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

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

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

Solving each part separately

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

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

tan (x) + 3 = 0

Move

3

to the right side

tan (x) + 3 = 0

Subtract

3

from both sides

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

Simplify

tan (x) = - 3

tan (x) = - 3

General solutions for

tan (x) = - 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{2 π}{3} + πn

x = \frac{2 π}{3} + πn

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

tan (x) - 3 = 0

Move

3

to the right side

tan (x) - 3 = 0

Add

3

to both sides

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

Simplify

tan (x) = 3

tan (x) = 3

General solutions for

tan (x) = 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}{3} + πn

x = \frac{π}{3} + πn

Combine all the solutions

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

Since the equation is undefined for:

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

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

Popular Examples

2sin(x)+3cot(x)-3csc(x)=0

2 sin (x) + 3 cot (x) - 3 csc (x) = 0

cos^2(θ)-3cos(θ)+2=0

cos^{2} (θ) - 3 cos (θ) + 2 = 0

cos(2x-pi/6)=-1/2 , pi/2 <= x<pi

cos (2 x - \frac{π}{6}) = - \frac{1}{2}, \frac{π}{2} \leq x < π

((sin(x)))/((4.2))=((sin(95)))/((10.26))

\frac{( sin ( x ) )}{( 4.2 )} = \frac{( sin ( 9 5 ^{\circ} ) )}{( 10.26 )}

-sin(x)=-cos(x)

- sin (x) = - cos (x)

Frequently Asked Questions (FAQ)

What is the general solution for cos(x)tan^2(x)-3cos(x)=0 ?
The general solution for cos(x)tan^2(x)-3cos(x)=0 is x=(2pi)/3+pin,x= pi/3+pin