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

The general solution for sqrt(2)tan(x)cos(x)=tan(x) is x=pin,x= pi/4+2pin,x=(7pi)/4+2pin

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Popular >

sqrt(2)tan(x)cos(x)=tan(x)

Solution

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

Solution

x = πn, x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Degrees

x = 0^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

tan (x)

from both sides

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

Factor

2 tan (x) cos (x) - tan (x) : tan (x) (2 cos (x) - 1)

2 tan (x) cos (x) - tan (x)

Factor out common term

tan (x)

= tan (x) (2 cos (x) - 1)

tan (x) (2 cos (x) - 1) = 0

Solving each part separately

tan (x) = 0 or 2 cos (x) - 1 = 0

tan (x) = 0 : x = πn

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

2 cos (x) - 1 = 0 : x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

2 cos (x) - 1 = 0

Move

1

to the right side

2 cos (x) - 1 = 0

Add

1

to both sides

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

Simplify

2 cos (x) = 1

2 cos (x) = 1

Divide both sides by

2

2 cos (x) = 1

Divide both sides by

2

\frac{2 cos ( x )}{2} = \frac{1}{2}

Simplify

\frac{2 cos ( x )}{2} = \frac{1}{2}

Simplify

\frac{2 cos ( x )}{2} : cos (x)

\frac{2 cos ( x )}{2}

Cancel the common factor:

2

= cos (x)

Simplify

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

cos (x) = \frac{2}{2}

cos (x) = \frac{2}{2}

cos (x) = \frac{2}{2}

General solutions for

cos (x) = \frac{2}{2}

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}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Combine all the solutions

x = πn, x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Popular Examples

sin(x)= 5/4

sin (x) = \frac{5}{4}

sin(2θ)=3cos(θ)

sin (2 θ) = 3 cos (θ)

solvefor x,tan(x)+|tan(x)|=1 solve for

x, tan (x) + ∣ tan (x) ∣ = 1

0=2sin^2(θ)+sqrt(3)sin(θ)

0 = 2 sin^{2} (θ) + 3 sin (θ)

solvefor y,x=arctan(x+y)solve for

y, x = arctan (x + y)

Frequently Asked Questions (FAQ)

What is the general solution for sqrt(2)tan(x)cos(x)=tan(x) ?
The general solution for sqrt(2)tan(x)cos(x)=tan(x) is x=pin,x= pi/4+2pin,x=(7pi)/4+2pin