What is the general solution for cos(θ)+2sec(θ)=-3 ?

The general solution for cos(θ)+2sec(θ)=-3 is θ=pi+2pin

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cos(θ)+2sec(θ)=-3

Solution

cos (θ) + 2 sec (θ) = - 3

Solution

θ = π + 2 πn

Degrees

θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

cos (θ) + 2 sec (θ) = - 3

Subtract

- 3

from both sides

cos (θ) + 2 sec (θ) + 3 = 0

Rewrite using trig identities

3 + cos (θ) + 2 sec (θ)

Use the basic trigonometric identity:

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

= 3 + \frac{1}{sec ( θ )} + 2 sec (θ)

3 + \frac{1}{sec ( θ )} + 2 sec (θ) = 0

Solve by substitution

3 + \frac{1}{sec ( θ )} + 2 sec (θ) = 0

Let:

sec (θ) = u

3 + \frac{1}{u} + 2 u = 0

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

3 + \frac{1}{u} + 2 u = 0

Multiply both sides by

u

3 + \frac{1}{u} + 2 u = 0

Multiply both sides by

u

3 u + \frac{1}{u} u + 2 uu = 0 \cdot u

Simplify

3 u + \frac{1}{u} u + 2 uu = 0 \cdot u

Simplify

\frac{1}{u} u : 1

\frac{1}{u} u

Multiply fractions:

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

= \frac{1 \cdot u}{u}

Cancel the common factor:

u

= 1

Simplify

2 uu : 2 u^{2}

2 uu

Apply exponent rule:

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

uu = u^{1 + 1}

= 2 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= 2 u^{2}

Simplify

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

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

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

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

Solve

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

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

Write in the standard form

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

2 u^{2} + 3 u + 1 = 0

Solve with the quadratic formula

2 u^{2} + 3 u + 1 = 0

Quadratic Equation Formula:

For

a = 2, b = 3, c = 1

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

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

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

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

Multiply the numbers:

4 \cdot 2 \cdot 1 = 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 2}

Separate the solutions

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

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

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

Add/Subtract the numbers:

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

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

Subtract the numbers:

- 3 - 1 = - 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

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

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

3 + \frac{1}{u} + 2 u

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

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

Substitute back

u = sec (θ)

sec (θ) = - \frac{1}{2}, sec (θ) = - 1

sec (θ) = - \frac{1}{2}, sec (θ) = - 1

sec (θ) = - \frac{1}{2} :

No Solution

sec (θ) = - \frac{1}{2}

sec (x) \leq - 1 or sec (x) \geq 1

No Solution

sec (θ) = - 1 : θ = π + 2 πn

sec (θ) = - 1

General solutions for

sec (θ) = - 1

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

θ = π + 2 πn

θ = π + 2 πn

Combine all the solutions

θ = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for cos(θ)+2sec(θ)=-3 ?
The general solution for cos(θ)+2sec(θ)=-3 is θ=pi+2pin