Question

The general solution of the trigonometric equation
$\cos \theta =-\frac{1}{\sqrt{2}}\text{is:}$

• A. $\theta =(2n+1)\pi \pm \frac{3\pi }{4}$, where $n\in Z$
• B. $\theta =n\pi \pm \frac{\pi }{4}$, where $n\in Z$
• C. $\theta =2n\pi \pm \frac{\pi }{4}$, where $n\in Z$
• D. $\theta =2n\pi \pm \frac{3\pi }{4}$, where $n\in Z$

Answer 1

The correct option is D $\theta =2n\pi \pm \frac{3\pi }{4}$, where $n\in Z$

Given that
$\cos \theta =-\frac{1}{\sqrt{2}}$
We know that $\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$
$\cos \theta$ is negative in 2nd and 3rd quadrant
$\Rightarrow \cos (\pi -\frac{\pi }{4})=\cos (\pi +\frac{\pi }{4})=-\frac{1}{\sqrt{2}}$
$\Rightarrow \cos \frac{3\pi }{4}=\cos \frac{5\pi }{4}=-\frac{1}{\sqrt{2}}$

Using $\cos \theta =\cos \alpha \Rightarrow \theta =2n\pi \pm \alpha$, where $n\in Z$

$\therefore$ The general solution of $\cos \theta =-\frac{1}{\sqrt{2}}$ is
$\theta =2n\pi \pm \frac{3\pi }{4}$ or $2n\pi \pm \frac{5\pi }{4}$, where $n\in Z$