Question

Find the general solution of \[\sqrt{3}\sec \theta =2\] .

Answer 1

Hint: Convert the expression in the problem in terms of $\sec \theta $ to $\cos \theta $, using the relation –

$\cos \theta =\dfrac{1}{\sec \theta }$ .

Complete step-by-step answer:


Now, compare the relation evaluated with $\cos x=\cos y$and use the general solution of $\cos x=\cos y$, which is given as –

$x=2n\pi \pm y$ .

Value of $\dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2}$ .


So, we have \[\sqrt{3}\sec \theta =2\]

On dividing the whole equation by $\sqrt{3}$ , we get –

\[\dfrac{\sqrt{3}\sec \theta }{\sqrt{3}}=\dfrac{2}{\sqrt{3}}\]

Or \[\sec \theta =\dfrac{2}{\sqrt{3}}\] ……………………………………….. (i)

Now, we can convert the above expression in terms of $\cos \theta $, by using the relation between $\cos \theta $and $\sec \theta $ , which is given as –

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

Hence, putting value of $\sec \theta $ from the equation (i) to the equation (ii), so, we can rewrite the equation (ii) as –

$\cos \theta =\dfrac{1}{\left( \dfrac{2}{\sqrt{3}} \right)}=\dfrac{\sqrt{3}}{2}\times 1$

$\cos \theta =\dfrac{\sqrt{3}}{2}$……………………………………. (iii)

Now, let us find the angle at which cosine function will give $\dfrac{\sqrt{3}}{2}$ .

So, we know –

$\cos \dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2}$ …………………………….. (iv)

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

$x=2n\pi \pm y$……………………………….. (vi)

Where, $x\in z$ i.e. n is an integer.

So, on comparing equation (v) i.e. $\cos \theta =\cos \dfrac{\pi }{6}$ with the equation $\cos x=\cos y$. And hence, we get the value of x and y as –

$x=\theta $ and $y=\dfrac{\pi }{6}$.

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 –

$\theta =2n\pi \pm \dfrac{\pi }{6}$ .

Where $n\in z$.

So, we get solution $\cos \dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2}$ by putting $n=0,1,2,3.........$ to

get the number of values of $\theta $ which will satisfy the given expression in the problem.


Note: One may use general solution for $\sin x=\sin y$ as well by writing $\cos \theta =\dfrac{\sqrt{3}}{2}$ in terms of $\sin \theta $ as –

$\sin \left( {{90}^{\circ }}-\theta \right)=\dfrac{\sqrt{3}}{2}=\sin \dfrac{\pi }{3}$ .

Or $\sin \left( \dfrac{\pi }{2}-\theta \right)=\sin \dfrac{\pi }{3}$.

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

$\begin{align}

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

 & \cos \theta =\cos \dfrac{\pi }{6} \\

 & \theta =2n\pi \pm \dfrac{\pi }{6}\text{.} \\

\end{align}$