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

The general solution for tan^2(x)cos(x)=tan^2(x) is x=pin

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

Solution

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

Solution

x = πn

Degrees

x = 0^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

tan^{2} (x)

from both sides

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

Factor

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

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

Factor out common term

tan^{2} (x)

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

tan^{2} (x) (cos (x) - 1) = 0

Solving each part separately

tan^{2} (x) = 0 or cos (x) - 1 = 0

tan^{2} (x) = 0 : x = πn

tan^{2} (x) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

tan (x) = 0

General solutions for

tan (x) = 0

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 = 0 + πn

x = 0 + πn

Solve

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

cos (x) - 1 = 0 : x = 2 πn

cos (x) - 1 = 0

Move

1

to the right side

cos (x) - 1 = 0

Add

1

to both sides

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

Simplify

cos (x) = 1

cos (x) = 1

General solutions for

cos (x) = 1

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 = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

Combine all the solutions

x = πn, x = 2 πn

Merge Overlapping Intervals

x = πn

Popular Examples

cos(x)sin(x)=3sin(x)

cos (x) sin (x) = 3 sin (x)

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

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

8sin(θ)-1=0

8 sin (θ) - 1 = 0

4sin(θ)cos(θ)=1

4 sin (θ) cos (θ) = 1

-64sin(8x)=0

- 64 sin (8 x) = 0

Frequently Asked Questions (FAQ)

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