Question
Find the general solution of : $\sec \theta+\sqrt{2}=0$

Answer

We have $\sec \theta+\sqrt{2}=0 \therefore \cos \theta=-\frac{1}{\sqrt{2}}$

$

\therefore \quad \cos \theta=\cos \frac{3 \pi}{4}\left(\text { As } \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}} \text { and } \cos (\pi-A)=-\cos A\right)

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The general solution of $\cos \theta=\cos \alpha$ is $\theta=2 n \pi \pm \alpha$, where $n \in Z$.

$\therefore \quad$ The general solution of $\cos \theta=\cos \frac{3 \pi}{4}$ is $\theta=2 n \pi \pm \frac{3 \pi}{4}$, where $n \in Z$.

$\therefore \quad$ The general solution of $\sec \theta=\sqrt{2}$ is $\theta=2 n \pi \pm \frac{3 \pi}{4}$, where $n \in Z$.

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