Question

Solve the inequality:
$\sin x+\sqrt{3}\cos x>0$

Answer 1

Hint: In this question, we first need to divide both the sides with 2. Then write the constant terms in terms of trigonometric ratios of some known angles of sine and cosine. Now, use the trigonometric ratios of compound angles formula and convert it to sine function. Then using the inequality we get the value of x.
$\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$

Complete step-by-step answer:
Now, from the given inequality in the question we have,
$\Rightarrow \sin x+\sqrt{3}\cos x>0$
Let us now divide with 2 on both the sides
$\Rightarrow \dfrac{1}{2}\sin x+\dfrac{\sqrt{3}}{2}\cos x>0$
As we already know that
$\cos \dfrac{\pi }{3}=\dfrac{1}{2},\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}$
Now, let us write the constant values in terms of standard values mentioned above
$\Rightarrow \sin x\cos \dfrac{\pi }{3}+\cos x\sin \dfrac{\pi }{3}>0$
As we already know that from the trigonometric ratios of compound angles formula that
$\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$
By, using this formula the above inequality can be converted as
$\Rightarrow \sin \left( x+\dfrac{\pi }{3} \right)>0$
As we already know that the sine function is positive in the first and second quadrants we can get the value of the angle where it lies,
Now, sine function is positive for the angles
$\Rightarrow 0<\theta <\pi$
Now, using this condition from the above inequality we get,
Now, on comparison we get that
$\theta =x+\dfrac{\pi }{3}$
Now, on applying the above mentioned condition we get,
$\Rightarrow 0 < x+ \dfrac{\pi }{3} < \pi$
Let us now subtract $\dfrac{\pi }{3}$ from each of the terms
$\Rightarrow 0-\dfrac{\pi }{3} < x < \pi -\dfrac{\pi }{3}$
Now, on further simplification we get,
$\therefore -\dfrac{\pi }{3} < x < \dfrac{2\pi }{3}$

Note:
Instead of writing in terms of sine function using the compound angles formula we can also convert it in terms of cosine function and simplify it further. Both the methods give the same result.
It is important to note that we need to consider all the possible quadrants in which the given inequality satisfies because considering only one of them changes the result completely.