Question

Write the set of values of a for which the equation $\sqrt{3}\sin x-\cos x=a$ has no solution.

Answer 1

Hint: The set of all the solutions of a given trigonometric equation constitute general solutions of the equation. In this question, you can use the sum formula of the sine trigonometric ratio of compound angles $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$

Complete step-by-step answer:

The equations that involve the trigonometric functions of a variable are called trigonometric equations. We will try to find the solutions of such equations. These equations have one or more trigonometric ratios of unknown angles. For example, $\cos x-\sin 2x=0$, is a trigonometric equation which does not satisfy all the values of x. Hence for such equations, we have to find the values of x or find the solution.

The given trigonometric equation is

$\sqrt{3}\sin x-\cos x=a$

Dividing both sides by 2, we get

$\dfrac{\sqrt{3}}{2}\sin x-\dfrac{1}{2}\cos x=\dfrac{a}{2}.......................(1)$

We know that, $\cos \left( \dfrac{\pi }{6} \right)=\dfrac{\sqrt{3}}{2},\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}$

Now put all the values in the equation (1), we get

$\cos \left( \dfrac{\pi }{6} \right)\sin x-\sin \left( \dfrac{\pi }{6} \right)\cos x=\dfrac{a}{2}$

Applying the sum formula of the sine trigonometric ratio of compound angles $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$ on the left side, we get

  $\sin \left( \dfrac{\pi }{6}-x \right)=\dfrac{a}{2}$

$\dfrac{\pi }{6}-x={{\sin }^{-1}}\left( \dfrac{a}{2} \right)$

Rearranging the terms, we get

$x={{\sin }^{-1}}\left( \dfrac{a}{2} \right)+\dfrac{\pi }{6}$

If a = 2 or a = -2, then the equation will possess a solution.

For no solution,

$a\in \left( -\infty ,-2 \right)\cup \left( 2,\infty \right)$

This is because the required set of the values has no solution.

Note: The rule of thumb with trigonometric equations is that there will be no solutions whenever $-1>\sin x,-1>\cos x,1<\sin x$ and $1<\cos x$.