Question

Find the general solution of $\cos \theta =-\dfrac{\sqrt{3}}{2}$.

Answer 1

Hint: Get the angle at which cosine function will give value of $-\dfrac{\sqrt{3}}{2}$ and use the following results:

$\to $ cosine function is negative in 2nd quadrant

$\to $$\cos \left( \pi -\theta \right)=-\cos \theta ,\cos \dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2}$
General solution of $\cos x=\cos y$ is given as $x=2n\pi \pm y,$ where $n\in z$. So, form the given expression in the problem as $\cos x=\cos y$and hence, put x, y to the equation mentioned above.

Complete step-by-step answer:

Here, we are given the trigonometric relation as

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

Now, let us find the angle with the cosine function, at which cosine will give $\dfrac{-\sqrt{3}}{2}$. As, we know the cosine function gives $\dfrac{\sqrt{3}}{2}$ at the angle $\dfrac{\pi }{6},{{30}^{\circ }}$ and we also know that cosine function is negative in 2nd and 3rd quadrant.

So, we can write trigonometric relation w.r.t the second quadrant as

$\cos \left( \pi -A \right)=-\cos A$ ………….(ii)

Now, as we know cosine function will give $\dfrac{\sqrt{3}}{2}\to {{30}^{\circ }},\dfrac{\pi }{6}$ so, let us put $A=\dfrac{\pi }{6}$ to equation (ii).

So, we get

$\cos \left( \pi -\dfrac{\pi }{4} \right)=-\cos \dfrac{\pi }{6},$

$\cos \left( \dfrac{5\pi }{6} \right)=-\dfrac{\sqrt{3}}{2}$………………(iii)

Hence, we can replace $-\dfrac{\sqrt{3}}{2}$ of the equation (i) $\to \cos \left( \dfrac{5\pi }{6} \right)$ with the help of above equation. So, we get

$\cos \theta =\cos \dfrac{5\pi }{6}$ ……………(iv)

Now, we know the general solution of trigonometric relation $\cos x=\cos y$, can be given as
$x=2n\pi \pm y$ ……………..(v)

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

So, on comparing equation (iv) i.e. $\cos \theta =\cos \dfrac{5\pi }{6}$, with the equation $\cos x=\cos y$, and hence, we get values of x and y as

$x=\theta ,y=\dfrac{5\pi }{6}$

Now, we can put the above values of x and y to the equation (v) to get the general solution of the given expression in the problem. So, we get

$\theta =2n\pi \pm \dfrac{5\pi }{6}$

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: Another approach to write the general solution would be given as

$\begin{align}

  & \cos \theta =-\dfrac{\sqrt{3}}{2} \\

 & -\sin \left( \dfrac{\sqrt{3}}{2}-\theta \right)=-\dfrac{\sqrt{3}}{2}, \\

 & \sin \left( \dfrac{\sqrt{3}}{2}-\theta \right)=\sin \dfrac{\pi }{2} \\

\end{align}$

Put general solution of $\sin x=\sin y$ i.e.

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

So, it can be another approach to get the number of solutions of the given expression.

Here, we can notice one more important fact that the representation of solution by different approach may differ but the solution will not differ on putting values of n to the general solution. So, don’t get confused with any different solution.

One can prove general solution of

$\cos x=\cos y$ as

$\cos x-\cos y=0$

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