What is the general solution for solvefor x,cos(2x)+2/(cos(2x))+3=0 ?

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

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Popular Trigonometry >

solvefor x,cos(2x)+2/(cos(2x))+3=0

Solution

solvefor

Solution

x = \frac{π}{2} + πn

Degrees

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

Solution steps

cos (2 x) + \frac{2}{cos ( 2 x )} + 3 = 0

Solve by substitution

cos (2 x) + \frac{2}{cos ( 2 x )} + 3 = 0

Let:

cos (2 x) = u

u + \frac{2}{u} + 3 = 0

u + \frac{2}{u} + 3 = 0 : u = - 1, u = - 2

u + \frac{2}{u} + 3 = 0

Multiply both sides by

u

u + \frac{2}{u} + 3 = 0

Multiply both sides by

u

uu + \frac{2}{u} u + 3 u = 0 \cdot u

Simplify

uu + \frac{2}{u} u + 3 u = 0 \cdot u

Simplify

uu : u^{2}

uu

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

uu = u^{1 + 1}

= u^{1 + 1}

Add the numbers:

1 + 1 = 2

= u^{2}

Simplify

\frac{2}{u} u : 2

\frac{2}{u} u

Multiply fractions:

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

= \frac{2 u}{u}

Cancel the common factor:

u

= 2

Simplify

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

u^{2} + 2 + 3 u = 0

u^{2} + 2 + 3 u = 0

u^{2} + 2 + 3 u = 0

Solve

u^{2} + 2 + 3 u = 0 : u = - 1, u = - 2

u^{2} + 2 + 3 u = 0

Write in the standard form

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

u^{2} + 3 u + 2 = 0

Solve with the quadratic formula

u^{2} + 3 u + 2 = 0

Quadratic Equation Formula:

For

a = 1, b = 3, c = 2

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot 2}{2 \cdot 1}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot 2}{2 \cdot 1}

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

3^{2} - 4 \cdot 1 \cdot 2

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

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

Add/Subtract the numbers:

- 3 + 1 = - 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

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

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

Subtract the numbers:

- 3 - 1 = - 4

= \frac{- 4}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 4}{2}

Apply the fraction rule:

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

= - \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= - 2

The solutions to the quadratic equation are:

u = - 1, u = - 2

u = - 1, u = - 2

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

u + \frac{2}{u} + 3

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = - 1, u = - 2

Substitute back

u = cos (2 x)

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

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

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

cos (2 x) = - 1

General solutions for

cos (2 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}

2 x = π + 2 πn

2 x = π + 2 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 = \frac{π}{2} + πn

cos (2 x) = - 2 :

No Solution

cos (2 x) = - 2

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = \frac{π}{2} + πn

Graph

Popular Examples

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Frequently Asked Questions (FAQ)

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