What is the general solution for 2sin(2x)-5tan(2x)=0 ?

The general solution for 2sin(2x)-5tan(2x)=0 is x=pin,x= pi/2+pin

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2sin(2x)-5tan(2x)=0

Solution

2 sin (2 x) - 5 tan (2 x) = 0

Solution

x = πn, x = \frac{π}{2} + πn

Degrees

x = 0^{\circ} + 18 0^{\circ} n, x = 9 0^{\circ} + 18 0^{\circ} n

Solution steps

2 sin (2 x) - 5 tan (2 x) = 0

Express with sin, cos

2 sin (2 x) - 5 tan (2 x)

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= 2 sin (2 x) - 5 \cdot \frac{sin ( 2 x )}{cos ( 2 x )}

Simplify

2 sin (2 x) - 5 \cdot \frac{sin ( 2 x )}{cos ( 2 x )} : \frac{2 sin ( 2 x ) cos ( 2 x ) - 5 sin ( 2 x )}{cos ( 2 x )}

2 sin (2 x) - 5 \cdot \frac{sin ( 2 x )}{cos ( 2 x )}

Multiply

5 \cdot \frac{sin ( 2 x )}{cos ( 2 x )} : \frac{5 sin ( 2 x )}{cos ( 2 x )}

5 \cdot \frac{sin ( 2 x )}{cos ( 2 x )}

Multiply fractions:

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

= \frac{sin ( 2 x ) \cdot 5}{cos ( 2 x )}

= 2 sin (2 x) - \frac{5 sin ( 2 x )}{cos ( 2 x )}

Convert element to fraction:

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

= \frac{2 sin ( 2 x ) cos ( 2 x )}{cos ( 2 x )} - \frac{sin ( 2 x ) \cdot 5}{cos ( 2 x )}

Since the denominators are equal, combine the fractions:

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

= \frac{2 sin ( 2 x ) cos ( 2 x ) - sin ( 2 x ) \cdot 5}{cos ( 2 x )}

= \frac{2 sin ( 2 x ) cos ( 2 x ) - 5 sin ( 2 x )}{cos ( 2 x )}

\frac{- 5 sin ( 2 x ) + 2 cos ( 2 x ) sin ( 2 x )}{cos ( 2 x )} = 0

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

- 5 sin (2 x) + 2 cos (2 x) sin (2 x) = 0

Factor

- 5 sin (2 x) + 2 cos (2 x) sin (2 x) : sin (2 x) (2 cos (2 x) - 5)

- 5 sin (2 x) + 2 cos (2 x) sin (2 x)

Factor out common term

sin (2 x)

= sin (2 x) (- 5 + 2 cos (2 x))

sin (2 x) (2 cos (2 x) - 5) = 0

Solving each part separately

sin (2 x) = 0 or 2 cos (2 x) - 5 = 0

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

sin (2 x) = 0

General solutions for

sin (2 x) = 0

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 x = 0 + 2 πn, 2 x = π + 2 πn

2 x = 0 + 2 πn, 2 x = π + 2 πn

Solve

2 x = 0 + 2 πn : x = πn

2 x = 0 + 2 πn

0 + 2 πn = 2 πn

2 x = 2 πn

Divide both sides by

2

2 x = 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 πn}{2}

Simplify

x = πn

x = πn

Solve

2 x = π + 2 πn : x = \frac{π}{2} + πn

2 x = π + 2 πn

Divide both sides by

2

2 x = π + 2 πn

Divide both sides by

2

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

Simplify

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = πn, x = \frac{π}{2} + πn

2 cos (2 x) - 5 = 0 :

No Solution

2 cos (2 x) - 5 = 0

Move

5

to the right side

2 cos (2 x) - 5 = 0

Add

5

to both sides

2 cos (2 x) - 5 + 5 = 0 + 5

Simplify

2 cos (2 x) = 5

2 cos (2 x) = 5

Divide both sides by

2

2 cos (2 x) = 5

Divide both sides by

2

\frac{2 cos ( 2 x )}{2} = \frac{5}{2}

Simplify

cos (2 x) = \frac{5}{2}

cos (2 x) = \frac{5}{2}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = πn, x = \frac{π}{2} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin(2x)-5tan(2x)=0 ?
The general solution for 2sin(2x)-5tan(2x)=0 is x=pin,x= pi/2+pin