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Question: How do you solve \( \sin x\cos x = \dfrac{1}{2} \) ?...

How do you solve sin⁡xcos⁡x=12\sin x\cos x = \dfrac{1}{2} ?

Explanation

Solution

In order to solve the above trigonometric equation, multiply both sides with the number 2 and use write 2sin⁡xcos⁡x2\sin x\cos x as sin⁡2x\sin 2x using the trigonometric identity . Take the inverse of sine on both sides and find the generalised solution of xx by using the fact that the period of sine function is 2π2\pi .

Complete step by step solution:
We are given a trigonometric equation sin⁡xcos⁡x=12\sin x\cos x = \dfrac{1}{2} .
sin⁡xcos⁡x=12\sin x\cos x = \dfrac{1}{2}
Multiplying both sides of the equation with the number 22 ,
2sin⁡xcos⁡x=12×2 2sin⁡xcos⁡x=1   2\sin x\cos x = \dfrac{1}{2} \times 2 \\\ 2\sin x\cos x = 1 \;
Using the identity of trigonometry of sine double angle i.e. sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x .So replacing 2sin⁡xcos⁡x2\sin x\cos x with sin⁡2x\sin 2x in the above equation , we have
sin⁡2x=1\sin 2x = 1
Now taking inverse of sine function on both sides of the equation
sin⁡−1(sin⁡2x)=sin⁡−1(1){\sin ^{ - 1}}\left( {\sin 2x} \right) = {\sin ^{ - 1}}\left( 1 \right)
Since sin⁡−1(sin⁡)=1{\sin ^{ - 1}}\left( {\sin } \right) = 1 as they both are inverse of each other
2x=sin⁡−1(1)2x = {\sin ^{ - 1}}\left( 1 \right)
sin⁡−1(1){\sin ^{ - 1}}\left( 1 \right) is nothing but an angle whose sine is equal to 1. As we know sin⁡π2=1→π2=sin⁡−1(1)\sin \dfrac{\pi }{2} = 1 \to \dfrac{\pi }{2} = {\sin ^{ - 1}}\left( 1 \right) . But if we try to generalise the solution, we have a period of sine function 2π2\pi as the sine function repeats itself after every 2π2\pi interval.
Hence,
2x=2nπ+π22x = 2n\pi + \dfrac{\pi }{2}
Dividing both side by 2 , we get

2x2=12(2nπ+π2) x=nπ+π4   \dfrac{{2x}}{2} = \dfrac{1}{2}\left( {2n\pi + \dfrac{\pi }{2}} \right) \\\ x = n\pi + \dfrac{\pi }{4} \;

Where n is an integer
Therefore, the solution to the given trigonometric equation is x=nπ+π4x = n\pi + \dfrac{\pi }{4}
So, the correct answer is “x=nπ+π4x = n\pi + \dfrac{\pi }{4}”.

Note : 1.One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.
2.The answer obtained should be in a generalised form.
3. The domain of sine function is in the interval [−π2,π2]\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] and the range is in the interval [−1,1]\left[ { - 1,1} \right] .