What is the general solution for tan(2x)+2cos(x)=0,0<= x<= 2pi ?

Q: What is the general solution for tan(2x)+2cos(x)=0,0<= x<= 2pi ?

The general solution for tan(2x)+2cos(x)=0,0<= x<= 2pi is x= pi/2 ,x=(3pi)/2 ,x=(7pi)/6 ,x=(11pi)/6

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tan(2x)+2cos(x)=0,0<= x<= 2pi

Solution

tan (2 x) + 2 cos (x) = 0, 0 \leq x \leq 2 π

Solution

x = \frac{π}{2}, x = \frac{3 π}{2}, x = \frac{7 π}{6}, x = \frac{11 π}{6}

Degrees

x = 9 0^{\circ}, x = 27 0^{\circ}, x = 21 0^{\circ}, x = 33 0^{\circ}

Solution steps

tan (2 x) + 2 cos (x) = 0, 0 \leq x \leq 2 π

Express with sin, cos

tan (2 x) + 2 cos (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

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

Rewrite using trig identities

sin (2 x) + 2 cos (2 x) cos (x)

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

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

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

Factor

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

2 cos (2 x) cos (x) + 2 cos (x) sin (x)

Rewrite as

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

Factor out common term

2 cos (x)

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

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

Solving each part separately

cos (x) = 0 or cos (2 x) + sin (x) = 0

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

cos (x) = 0, 0 \leq x \leq 2 π

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

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{2}, x = \frac{3 π}{2}

cos (2 x) + sin (x) = 0, 0 \leq x \leq 2 π : x = \frac{7 π}{6}, x = \frac{11 π}{6}, x = \frac{π}{2}

cos (2 x) + sin (x) = 0, 0 \leq x \leq 2 π

Rewrite using trig identities

cos (2 x) + sin (x)

Use the Double Angle identity:

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

= 1 - 2 sin^{2} (x) + sin (x)

1 + sin (x) - 2 sin^{2} (x) = 0

Solve by substitution

1 + sin (x) - 2 sin^{2} (x) = 0

Let:

sin (x) = u

1 + u - 2 u^{2} = 0

1 + u - 2 u^{2} = 0 : u = - \frac{1}{2}, u = 1

1 + u - 2 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- 2 u^{2} + u + 1 = 0

Solve with the quadratic formula

- 2 u^{2} + u + 1 = 0

Quadratic Equation Formula:

For

a = - 2, b = 1, c = 1

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 2 ) \cdot 1}{2 ( - 2 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 2 ) \cdot 1}{2 ( - 2 )}

1^{2} - 4 (- 2) \cdot 1 = 3

1^{2} - 4 (- 2) \cdot 1

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 2) \cdot 1

Apply rule

- (- a) = a

= 1 + 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- 1 \pm 3}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- 1 + 3}{2 ( - 2 )}, u_{2} = \frac{- 1 - 3}{2 ( - 2 )}

u = \frac{- 1 + 3}{2 ( - 2 )} : - \frac{1}{2}

\frac{- 1 + 3}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 3}{- 2 \cdot 2}

Add/Subtract the numbers:

- 1 + 3 = 2

= \frac{2}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{- 4}

Apply the fraction rule:

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

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

u = \frac{- 1 - 3}{2 ( - 2 )} : 1

\frac{- 1 - 3}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 3}{- 2 \cdot 2}

Subtract the numbers:

- 1 - 3 = - 4

= \frac{- 4}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{- 4}

Apply the fraction rule:

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

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = - \frac{1}{2}, u = 1

Substitute back

u = sin (x)

sin (x) = - \frac{1}{2}, sin (x) = 1

sin (x) = - \frac{1}{2}, sin (x) = 1

sin (x) = - \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{7 π}{6}, x = \frac{11 π}{6}

sin (x) = - \frac{1}{2}, 0 \leq x \leq 2 π

General solutions for

sin (x) = - \frac{1}{2}

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}

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{7 π}{6}, x = \frac{11 π}{6}

sin (x) = 1, 0 \leq x \leq 2 π : x = \frac{π}{2}

sin (x) = 1, 0 \leq x \leq 2 π

General solutions for

sin (x) = 1

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}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{2}

Combine all the solutions

x = \frac{7 π}{6}, x = \frac{11 π}{6}, x = \frac{π}{2}

Combine all the solutions

x = \frac{π}{2}, x = \frac{3 π}{2}, x = \frac{7 π}{6}, x = \frac{11 π}{6}

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Frequently Asked Questions (FAQ)

What is the general solution for tan(2x)+2cos(x)=0,0<= x<= 2pi ?
The general solution for tan(2x)+2cos(x)=0,0<= x<= 2pi is x= pi/2 ,x=(3pi)/2 ,x=(7pi)/6 ,x=(11pi)/6