Question

How do you solve $\sin x=\dfrac{\sqrt{3}}{2}$? \[\]

Answer 1

Hint: We recall trigonometric equations, principal solution and general solution of a trigonometric equation. . We use the standard solutions of the trigonometric equation $\sin \theta =\sin \alpha $ that is $\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha $ where $n$ is any arbitrary integer and $\alpha $ is the principal solution. \[\]

Complete step by step answer:
We know that a trigonometric equation is an equation with trigonometric functions with unknown arguments as measure of angles. When we are asked to solve a trigonometric equation we have to find all possible measures of unknown angles.
We know that the first solution of the trigonometric equation within the interval $\left[ 0,2\pi \right]$is called principal solution and using periodicity all possible solutions obtained with integer $n$ are called general solutions. The general solution of the trigonometric equation $\sin \theta =\sin \alpha $ with principal solution $\theta =\alpha $ are given by
\[\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha \]
Here $n$ is any integer may be negative, positive or zero. We are given the following trigonometric equation in the question
\[\sin x=\dfrac{\sqrt{3}}{2}\]
We know from basic trigonometric table that $\sin {{60}^{\circ }}=\sin \left( \dfrac{\pi }{3} \right)=\dfrac{\sqrt{3}}{2}$. So we can write
\[\sin x=\sin \left( \dfrac{\pi }{3} \right)\]

We see here that $x=\dfrac{\pi }{3}$ is the principal solution since it is the first $x\in \left[ 0,2\pi \right)$ that satisfies given sine trigonometric equations . The general solutions of the given equation are
\[x=n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{3}\]
We can take different values of $n=...-2,-1,-,1,2...$and have the solutions as
\[x=...\dfrac{-5\pi }{3},\dfrac{-4\pi }{3},\dfrac{\pi }{3},\dfrac{2\pi }{3},\dfrac{7\pi }{3},...\]

Note: We must not confuse between the solutions of $\sin \theta =\sin \alpha $ and $\tan \theta =\tan \alpha $ whose solutions are given by $\theta =n\pi +\alpha $. We note that for both $\sin \theta =\sin \alpha ,\tan \theta =\tan \alpha $ the principal angel always lie in the interval $\left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)$ but for $\cos \theta =\cos \alpha $ the principal angle $\theta $ lies in the interval $\left[ 0,\pi \right)$ which has solutions $\theta =2n\pi \pm \alpha $. We should also remember the special cases of $\sin \theta =\sin \alpha $ which are $\sin \theta =0,\sin \theta =1,\sin \theta =-1$. The solutions of $\sin \theta =0$ are integral multiple of $\pi $ which means $\theta =n\pi $. The solutions of $\sin \theta =1$ are given by $\theta =\dfrac{4n+1}{2}\pi $ and the solutions of $\sin \theta =-1$ are given by $\theta =\dfrac{4n+3}{2}\pi $.