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Popular Trigonometry >

1/(cos(θ))=2cos(θ)

Solution

\frac{1}{cos ( θ )} = 2 cos (θ)

Solution

θ = 0.78539 \dots + 2 πn, θ = 2 π - 0.78539 \dots + 2 πn, θ = 2.35619 \dots + 2 πn, θ = - 2.35619 \dots + 2 πn

+1

Degrees

θ = 4 5^{\circ} + 36 0^{\circ} n, θ = 31 5^{\circ} + 36 0^{\circ} n, θ = 13 5^{\circ} + 36 0^{\circ} n, θ = - 13 5^{\circ} + 36 0^{\circ} n

Solution steps

\frac{1}{cos ( θ )} = 2 cos (θ)

Solve by substitution

\frac{1}{cos ( θ )} = 2 cos (θ)

Let:

cos (θ) = u

\frac{1}{u} = 2 u

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

\frac{1}{u} = 2 u

Multiply both sides by

u

\frac{1}{u} = 2 u

Multiply both sides by

u

\frac{1}{u} u = 2 uu

Simplify

2 uu : 2 u^{2}

\frac{1}{u} u = 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}

1 = 2 u^{2}

1 = 2 u^{2}

Solve

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

1 = 2 u^{2}

Switch sides

2 u^{2} = 1

Divide both sides by

2

2 u^{2} = 1

Divide both sides by

2

\frac{2 u ^{2}}{2} = \frac{1}{2}

Simplify

u^{2} = \frac{1}{2}

u^{2} = \frac{1}{2}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

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

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

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

\frac{1}{u}

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

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

Substitute back

u = cos (θ)

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

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

cos (θ) = \frac{1}{2} : θ = arccos (\frac{1}{2}) + 2 πn, θ = 2 π - arccos (\frac{1}{2}) + 2 πn

cos (θ) = \frac{1}{2}

Apply trig inverse properties

cos (θ) = \frac{1}{2}

General solutions for

cos (θ) = \frac{1}{2}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

θ = arccos (\frac{1}{2}) + 2 πn, θ = 2 π - arccos (\frac{1}{2}) + 2 πn

θ = arccos (\frac{1}{2}) + 2 πn, θ = 2 π - arccos (\frac{1}{2}) + 2 πn

cos (θ) = - \frac{1}{2} : θ = arccos (- \frac{1}{2}) + 2 πn, θ = - arccos (- \frac{1}{2}) + 2 πn

cos (θ) = - \frac{1}{2}

Apply trig inverse properties

cos (θ) = - \frac{1}{2}

General solutions for

cos (θ) = - \frac{1}{2}

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

θ = arccos (- \frac{1}{2}) + 2 πn, θ = - arccos (- \frac{1}{2}) + 2 πn

θ = arccos (- \frac{1}{2}) + 2 πn, θ = - arccos (- \frac{1}{2}) + 2 πn

Combine all the solutions

θ = arccos (\frac{1}{2}) + 2 πn, θ = 2 π - arccos (\frac{1}{2}) + 2 πn, θ = arccos (- \frac{1}{2}) + 2 πn, θ = - arccos (- \frac{1}{2}) + 2 πn

Show solutions in decimal form

θ = 0.78539 \dots + 2 πn, θ = 2 π - 0.78539 \dots + 2 πn, θ = 2.35619 \dots + 2 πn, θ = - 2.35619 \dots + 2 πn

Graph

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