Question

Find the general solution of $\sec \theta =\sqrt{2}$.

Answer 1

Hint: Use the relation between $\sec \theta ,\cos \theta \to \cos \theta =\dfrac{1}{\sec \theta }$, to get the given equation in terms of $\cos \theta $. Value of $\cos \dfrac{\pi }{4}\to \dfrac{1}{\sqrt{2}}$ and general solution of $\cos x=\cos y$ is given as

$x=2n\pi \pm y$

Hence, use the above results to get the solution.

Complete step-by-step answer:

Given expression in the problem is

$\sec \theta =\sqrt{2}$ ……………(i)

Now, as we know the relation between $\sec \theta ,\cos \theta $ can be given as

$\cos \theta =\dfrac{1}{\sec \theta }$ …………….(ii)

So, we can get the equation (i) in more familiar form i.e. in terms of $\cos \theta $, with the help of equation (ii) as

$\cos \theta =\dfrac{1}{\sqrt{2}}$ ……………..(iii)

Now, let us find the angle at which the cosine function will give $\dfrac{1}{\sqrt{2}}$ .

So, we know $\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$ ………………….(iv)

Hence, we can replace the right hand side of the equation (iii)

 $\begin{align}

  & \to \dfrac{1}{\sqrt{2}}, \\

 & \to cos\dfrac{\pi }{4} \\

\end{align}$

With the help of equation (iv). So, we can rewrite the equation (iii) as

$\cos \theta =\cos \dfrac{\pi }{4}$ ………………….(v)

Now, we know the general solution of trigonometric relation $\cos x=\cos y$, can be given as

$x=2n\pi \pm y$ …………………(vi)

Where, $n\in z\to n$ is an integer.

So, on comparing equation (v) $\to \cos \theta =\cos \left( \dfrac{\pi }{4} \right)$, with the

equation $\operatorname{cosx}=cosy$, and hence, we get values of x and y as

$x=\theta ,y=\dfrac{\pi }{4}$

Now, we can put the above values of x and y to the equation (vi) to get the general solution

of the given expression in the problem. So, we get

$\theta =2n\pi \pm \dfrac{\pi }{4}$

Where $n\in z$

So, we put n = 0, 1, 2, 3……….. to get the number of values of $\theta $ which will satisfy the given expression in the problem.

Note: One may convert the given relation as

$\begin{align}

  & \cos \theta =\dfrac{1}{\sqrt{2}} \\

 & \sin \left( 90-\theta \right)=\dfrac{1}{\sqrt{2}}=\sin \dfrac{\pi }{4}, \\

 & \sin \left( \dfrac{\pi }{2}-\theta \right)=\sin \dfrac{\pi }{4} \\

\end{align}$

Use general solution of $\sin x=\sin y$, given as

$x=n\pi +{{\left( -1 \right)}^{n}}y$

So, it can be another way and another representation of the solution of the same equation.

One can prove general solution of

$\cos x=\cos y\to \cos x-\cos y=0$

Use: $\operatorname{cosC}-\cos D=-2\sin \dfrac{C-D}{2}\sin \dfrac{C+D}{2}$

So, we get

$-2\sin \left( \dfrac{x-y}{2} \right)\sin \dfrac{x+y}{2}=0$

So, put $\sin \left( \dfrac{x-y}{2} \right)=0,\sin \left( \dfrac{x+y}{2} \right)=0$

And use the general solution of equation $\sin \theta -0\to \theta =n\pi $ to get the general solution of $\cos x=\cos y.$