Question

How do you solve \[2sinx-1=0~\] over the interval 0 to $2\pi$?

Answer 1

Hint: Solve the given equation for ‘\[sinx\]’. Replace the obtained value of \[sinx\] with a primary sine angle (say$'\sin \alpha '$). Find the general solution of ‘x’ by the formula $x=n\pi +{{\left( -1 \right)}^{n}}\alpha $(where n= 0, 1, 2, 3….). Obtain the value of ‘x’ for a different value of n. Take the values that lie between the interval 0 to 2pi to get the required solution.

Complete step by step solution:
According to the question,
\[2sinx-1=0~\]
Adding ‘1’ in both sides of the equation, we have
\[\Rightarrow 2sinx=1\]
Multiplying both sides by ‘$\dfrac{1}{2}$’, we get
\[\Rightarrow sinx=\dfrac{1}{2}\]
We know that, $\sin \dfrac{\pi }{6}=\dfrac{1}{2}$
From the above two equations;
$\Rightarrow \sin x=\sin \dfrac{\pi }{6}$
General solution of $\sin x=\sin \alpha $ is,
$x=n\pi +{{\left( -1 \right)}^{n}}\alpha $
Referring to this question,
$x=n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{6}$ (where n =0, 1, 2, 3…..)
Putting the value of n= 0, 1, 2, 3….. in the above equation we get,
$x=\dfrac{\pi }{6},\dfrac{5\pi }{6},\dfrac{13\pi }{6},\dfrac{17\pi }{6}.....$
From these obtained values of ‘x’ only $\dfrac{\pi }{6},\dfrac{5\pi }{6}$are present in the interval of 0 to 2pi$\Rightarrow x=\dfrac{\pi }{6}\text{ or }\dfrac{\text{5}\pi }{6}$ (Where, $\dfrac{\pi }{6}$ present in 1st quadrant and $\dfrac{\text{5}\pi }{6}$ is present in 2nd quadrant)
This is the required solution of the given question.

Note: First the general solution of \[sinx\] should be found using the formula $x=n\pi +{{\left( -1 \right)}^{n}}\alpha $. Then for this particular question $\alpha =\dfrac{\pi }{6}$ and n=0,1 as the interval given in the question is 0 to 2pi. So only two values of ‘x’ i.e. $\dfrac{\pi }{6}$ and $\dfrac{\text{5}\pi }{6}$ should lie in the given interval.