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Question: How do you find the exact solution of the equation \[4\sin x\cos x=1\] in the interval \[\left[ 0,2\...

How do you find the exact solution of the equation 4sin⁡xcos⁡x=14\sin x\cos x=1 in the interval [0,2π)?\left[ 0,2\pi \right)?

Explanation

Solution

We are asked to find the exact solution of 4sin⁡xcos⁡x=14\sin x\cos x=1 in the interval [0,2π).\left[ 0,2\pi \right). We will first learn what are solutions and then we will use sin 2x = 2 sin x cos x property to simplify our given equation. Once we have that we will simplify our equation and we will find the solution. We will also use that sin is periodic so it can have many solutions. The best way is we just have to find the solutions which are in [0,2π).\left[ 0,2\pi \right).

Complete step by step answer:
We are given 4sin⁡xcos⁡x=14\sin x\cos x=1 and we have to find the solution. The solutions are those values of x that will satisfy the given equation. So, we have to find the value of x which will satisfy our equation 4 sin x cos x = 1. Now to simplify and solve this problem we start by simplifying the equation 4 sin x cos x = 1. We will try to convert this into the function of any trigonometric ratios. So, as we know 2 sin x cos x = sin 2x, so we use this. So, we get,
4sin⁡xcos⁡x=14\sin x\cos x=1
⇒2(2sin⁡xcos⁡x)=1[As 2×2=4]\Rightarrow 2\left( 2\sin x\cos x \right)=1\left[ \text{As }2\times 2=4 \right]
Now using 2 sin x cos x = 1, we get,
⇒2sin⁡2x=1\Rightarrow 2\sin 2x=1
Now, we will divide both the sides by 2, so we get,
⇒2sin⁡2x2=12\Rightarrow \dfrac{2\sin 2x}{2}=\dfrac{1}{2}
So, we have
⇒sin⁡2x=12\Rightarrow \sin 2x=\dfrac{1}{2}
Now, we know that sin⁡θ\sin \theta is positive in quadrant I and quadrant II. So, for quadrant I,
sin⁡2x=12\sin 2x=\dfrac{1}{2}
As sin⁡(π6)=12\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2} so
So, sin⁡2x=sin⁡π6\sin 2x=\sin \dfrac{\pi }{6}
Comparing, we get,
⇒2x=π6\Rightarrow 2x=\dfrac{\pi }{6}
Since sin is periodic, so the solution in quadrant I is given as x=π12+2nπ2x=\dfrac{\pi }{12}+\dfrac{2n\pi }{2} where n = 0, 1, 2, 3, ...
When we put n = 0, we get, x=π12∈[0,2π)x=\dfrac{\pi }{12}\in \left[ 0,2\pi \right)
When we put n = 1, we get,
x=π12+2π2x=\dfrac{\pi }{12}+\dfrac{2\pi }{2}
⇒x=13π2∈[0,2π)\Rightarrow x=\dfrac{13\pi }{2}\in \left[ 0,2\pi \right)
If we find more then they will go out of [0,2π),\left[ 0,2\pi \right), so, from here we get x=π12x=\dfrac{\pi }{12} and x=13π12.x=\dfrac{13\pi }{12}.
Now for quadrant II, we get,
sin⁡(2x)=12\sin \left( 2x \right)=\dfrac{1}{2}
In quadrant II, sin⁡(π6)=12.\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}. So, sin⁡(2x)=sin⁡(5π6)\sin \left( 2x \right)=\sin \left( \dfrac{5\pi }{6} \right)
Comparing, we get,
⇒2x=5π6\Rightarrow 2x=\dfrac{5\pi }{6}
The solution in quadrant II will be given as 2x=5π6+2nπ2x=\dfrac{5\pi }{6}+2n\pi where n = 0, 1, 2, …
So, x=5π12+πn.x=\dfrac{5\pi }{12}+\pi n.
When we put n = 0, we get, x=5π12∈[0,2π).x=\dfrac{5\pi }{12}\in \left[ 0,2\pi \right).
When we put n = 1, we get, x=5π12+π=17π12∈[0,2π).x=\dfrac{5\pi }{12}+\pi =\dfrac{17\pi }{12}\in \left[ 0,2\pi \right).
If we move further we go out at [0,2π).\left[ 0,2\pi \right). So from here we get 2 solutions that is n=5π12n=\dfrac{5\pi }{12} and n=17π12.n=\dfrac{17\pi }{12}.
So, we get the exact solution of 4 sin x cos x = 1 on [0,2π)\left[ 0,2\pi \right) are π12,5π12,13π12,17π12.\dfrac{\pi }{12},\dfrac{5\pi }{12},\dfrac{13\pi }{12},\dfrac{17\pi }{12}.

Note: While finding all the solutions, we need to know that we just don’t need the solution in quadrant but we also have to find the value of the solution that may lie in other quadrants. Another way to find the solution is to find the graph of y = sin 2x and y = 1, then the point where the graph is on [0,2π)\left[ 0,2\pi \right) are the required solution.