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Question: How do you find all \(\sin 6x + \sin 2x = 0\) in the interval \([0,2\pi )\) ?...

How do you find all sin⁡6x+sin⁡2x=0\sin 6x + \sin 2x = 0 in the interval [0,2π)[0,2\pi ) ?

Explanation

Solution

In trigonometric functions, we insert the value of an angle as input and get a numerical value as the output, that is, the trigonometric functions convert the value of angles into numerical values. They are related to each other by several identities, we will use one such identity to solve this question.
The identity that we will use states that the sum of the sine of an angle A and side of another angle B is equal to twice the product of the sine of the half of sum of angle A and B and the cosine of the half of the difference of angle A and B, that is, sin⁡A+sin⁡B=2sin⁡(A+B2)cos⁡(A−B2)\sin A + \sin B = 2\sin (\dfrac{{A + B}}{2})\cos (\dfrac{{A - B}}{2}) .

Complete step by step answer:
We are given that sin⁡6x+sin⁡2x=0\sin 6x + \sin 2x = 0
Using the identity sin⁡A+sin⁡B=2sin⁡(A+B2)cos⁡(A−B2)\sin A + \sin B = 2\sin (\dfrac{{A + B}}{2})\cos (\dfrac{{A - B}}{2})
We get –
sin⁡6x+sin⁡2x=2sin⁡6x+2x2cos⁡6x−2x2 ⇒sin⁡6x+sin⁡2x=2sin⁡4xcos⁡2x  \sin 6x + \sin 2x = 2\sin \dfrac{{6x + 2x}}{2}\cos \dfrac{{6x - 2x}}{2} \\\ \Rightarrow \sin 6x + \sin 2x = 2\sin 4x\cos 2x \\\
Putting this value in the given equation, we get –
2sin⁡4xcos⁡2x=02\sin 4x\cos 2x = 0
As the product of 2, sin⁡4x and cos⁡2x2,\,\sin 4x\,and\,\cos 2x is zero, so any of these functions is zero. As
2≠02 \ne 0 , we get –
sin⁡4x=0 or cos⁡2x=0\sin 4x = 0\,or\,\cos 2x = 0
Now, in the interval [0,2π)[0,2\pi ) , the sine of a function is zero at 0, π and2π0,\,\pi \,and2\pi and cosine of a function is zero at π2 and 3π2\dfrac{\pi }{2}\,and\,\dfrac{{3\pi }}{2} , so we get –
4x=0, 4x=π, or 4x=2π ⇒x=0, x=π4, or x=π2  4x = 0,\,4x = \pi ,\,or\,4x = 2\pi \\\ \Rightarrow x = 0,\,x = \dfrac{\pi }{4},\,or\,x = \dfrac{\pi }{2} \\\
And
2x=π2 or 2x=3π2 ⇒x=π4 or x=3π4  2x = \dfrac{\pi }{2}\,or\,2x = \dfrac{{3\pi }}{2} \\\ \Rightarrow x = \dfrac{\pi }{4}\,or\,x = \dfrac{{3\pi }}{4} \\\
Thus x=0, π4, π2, 3π4x = 0,\,\dfrac{\pi }{4},\,\dfrac{\pi }{2},\,\dfrac{{3\pi }}{4}
Hence when sin⁡6x+sin⁡2x=0\sin 6x + \sin 2x = 0 , we get x=0, π4, π2, 3π4x = 0,\,\dfrac{\pi }{4},\,\dfrac{\pi }{2},\,\dfrac{{3\pi }}{4} .

Note: Trigonometric functions are periodic functions, that is, the output value of the trigonometric functions repeats itself after a specific interval of input values. The interval [0,2π][0,2\pi ] is the principal interval and the values repeat after this interval. But the signs of different trigonometric functions are different in different quadrants, so the sign of the function doesn’t repeat.