# Find all solutions of the equation. \sec(\theta)(2\cos(\theta) - \sqrt{2}) = 0

## Question:

Find all solutions of the equation.

{eq}\displaystyle\; \sec(\theta)\left(2\cos(\theta) - \sqrt{2}\right) = 0 {/eq}

## Trigonometric Equations:

A trigonometric equation is a mathematical equation that has atleast one of the trigonometric functions such as {eq}\text{sine(sin), tangent(tan), cosine(cos), cosecant(csc), secant(sec) or cotangent(cot)} {/eq}.

We have the equation:

• {eq}\displaystyle\; \sec(\theta)\left(2\cos(\theta) - \sqrt{2}\right) = 0 {/eq}

Equating each factor to zero, we get:

• {eq}\sec(\theta) = 0 {/eq}................................................................................(i)
• {eq}\left(2\cos(\theta) - \sqrt{2}\right) = 0 {/eq}................................................................................(ii)

From equation (i):

• {eq}\sec(\theta) = \dfrac{1}{\cos (\theta)} = 0 {/eq}

Secant ranges between {eq}y\leq -1 \text{ and } y\geq 1 {/eq}. Since{eq}0 {/eq} does not lie in this range, then there is no solution for this.

From equation (ii):

• {eq}\left(2\cos(\theta) - \sqrt{2}\right) = 0 {/eq}

Adding {eq}\sqrt{2} {/eq} to both sides of the equation:

• {eq}2\cos(\theta) = \sqrt{2} {/eq}

Dividing both sides of the equation by {eq}2 {/eq}:

• {eq}\cos(\theta) = \dfrac{\sqrt{2}}{2} {/eq}

Taking the inverse of cosine on both sides of the equation, we get:

• {eq}\theta = \cos^{-1} \dfrac{\sqrt{2}}{2} {/eq}
• {eq}\theta =45^\circ {/eq}

Since cosine is positive on the first and on the fourth quadrant, then:

• {eq}\theta =\left(45^\circ, \quad 315^\circ\right) {/eq}

Cosine has a range of {eq}360^\circ \text{ or } 2\pi {/eq}. Therefore:

• {eq}\theta =\boxed{\color{blue}{\left(45^\circ + 360k, \quad 315^\circ + 360k\right)}} {/eq}

Or:

• {eq}\theta =\boxed{\color{blue}{\left(\dfrac{\pi}{4}+ 2\pi k, \quad \dfrac{7\pi}{4} + 2\pi k\right)}} {/eq}

Where {eq}k {/eq} is any integer.