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Question: How do you solve \(4{{\sin }^{2}}x=1\) for \(x\) in the interval \(\left[ 0,2\pi \right)\)?...

How do you solve 4sin⁡2x=14{{\sin }^{2}}x=1 for xx in the interval [0,2π)\left[ 0,2\pi \right)?

Explanation

Solution

We have been given a quadratic equation of sin⁡x\sin x. We divide both sides of the equation by the constant 4. Then we take the square root on both sides of the equation. From that we find the exact solutions for the equation 4sin⁡2x=14{{\sin }^{2}}x=1 for xx in the interval [0,2π)\left[ 0,2\pi \right).

Complete step by step answer:
The given equation of sin⁡x\sin x is 4sin⁡2x=14{{\sin }^{2}}x=1.
We divide both sides of the equation by the constant 4 and get
4sin⁡2x4=14 ⇒sin⁡2x=14 \begin{aligned} & \dfrac{4{{\sin }^{2}}x}{4}=\dfrac{1}{4} \\\ & \Rightarrow {{\sin }^{2}}x=\dfrac{1}{4} \\\ \end{aligned}
We take square roots on both sides of the equation. As the equation is a quadratic one, the number of roots will be 2 and they are equal in value but opposite in sign.
sin⁡2x=14 ⇒sin⁡x=±12 \begin{aligned} & \sqrt{{{\sin }^{2}}x}=\sqrt{\dfrac{1}{4}} \\\ & \Rightarrow \sin x=\pm \dfrac{1}{2} \\\ \end{aligned}
We know that in the principal domain or the periodic value of −π2≤x≤π2-\dfrac{\pi }{2}\le x\le \dfrac{\pi }{2} for sin⁡x\sin x, if we get sin⁡a=sin⁡b\sin a=\sin b where −π2≤a,b≤π2-\dfrac{\pi }{2}\le a,b\le \dfrac{\pi }{2} then a=ba=b.
We have sin⁡x=12\sin x=\dfrac{1}{2}, the value of sin⁡(π6),sin⁡(5π6)\sin \left( \dfrac{\pi }{6} \right),\sin \left( \dfrac{5\pi }{6} \right) as 12\dfrac{1}{2} in the domain of [0,2π)\left[ 0,2\pi \right).
We have sin⁡x=−12\sin x=-\dfrac{1}{2}, the value of sin⁡(7π6),sin⁡(11π6)\sin \left( \dfrac{7\pi }{6} \right),\sin \left( \dfrac{11\pi }{6} \right) as −12-\dfrac{1}{2} in the domain of [0,2π)\left[ 0,2\pi \right).
Therefore, sin⁡x=±12\sin x=\pm \dfrac{1}{2} gives x=π6,5π6,7π6,11π6x=\dfrac{\pi }{6},\dfrac{5\pi }{6},\dfrac{7\pi }{6},\dfrac{11\pi }{6} as primary value.

Note:
Although for elementary knowledge the principal domain is enough to solve the problem. But if mentioned to find the general solution then the domain changes to −∞≤x≤∞-\infty \le x\le \infty . In that case we have to use the formula x=nπ+(−1)nax=n\pi +{{\left( -1 \right)}^{n}}a for sin⁡(x)=sin⁡a\sin \left( x \right)=\sin a where −π2≤a≤π2-\dfrac{\pi }{2}\le a\le \dfrac{\pi }{2}. For our given problem sin⁡x=±12\sin x=\pm \dfrac{1}{2}, the primary solution is x=±π6x=\pm \dfrac{\pi }{6}.
The general solution will be x=(nπ±(−1)nπ6)x=\left( n\pi \pm {{\left( -1 \right)}^{n}}\dfrac{\pi }{6} \right). Here n∈Zn\in \mathbb{Z}.