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Question: How do you solve \({\sin ^2}x + \sin x = 0\) and find all the solutions in the interval \([0,2\pi )\...

How do you solve sin⁡2x+sin⁡x=0{\sin ^2}x + \sin x = 0 and find all the solutions in the interval [0,2π)[0,2\pi )?

Explanation

Solution

Here in this question, we have to find the value of x at which the given equation sin⁡2x+sin⁡x=0{\sin ^2}x + \sin x = 0 satisfy. The question is based on the topic of trigonometry. By using the table of trigonometry ratios of standard angle we determine the value of x.

Complete step by step answer:
The given question is based on the topic of trigonometry. The sine is one of the trigonometry ratios. Here we have to find the values of x and that values of x should satisfy the given equation and the value of x should be in between 0 and 2π2\pi . Here the interval is represented as [ ) this indicates the closed open interval.
Now consider the given equation
sin⁡2x+sin⁡x=0{\sin ^2}x + \sin x = 0
The sin⁡2x{\sin ^2}x can be expressed in the terms of product so it is written as
⇒sin⁡x.sin⁡x+sin⁡x=0\Rightarrow \sin x.\sin x + \sin x = 0
Now we can take sin⁡x\sin x as a common in the LHS of the above equation. It is written as
⇒sin⁡x(sin⁡x+1)=0\Rightarrow \sin x(\sin x + 1) = 0
So, we have
⇒sin⁡x=0\Rightarrow \sin x = 0 or sin⁡x+1=0\sin x + 1 = 0
Let we consider sin⁡x=0\sin x = 0
Now we have to find the value of x at where sin⁡x=0\sin x = 0
Therefore ⇒x=sin⁡−1(0) \Rightarrow x = {\sin ^{ - 1}}(0)
So, we have ⇒x=±nπ\Rightarrow x = \pm n\pi
Here we have to find the solutions in the interval [0,2π)[0,2\pi ). Therefore x=0,πx = 0,\pi .
We can’t take 2π2\pi , because it doesn’t contain the interval. Now consider sin⁡x+1=0\sin x + 1 = 0. This can be written as sin⁡x=−1\sin x = - 1
Now we have to find the value of x at where sin⁡x=−1\sin x = - 1
Therefore ⇒x=sin⁡−1(−1) \Rightarrow x = {\sin ^{ - 1}}( - 1)
So, we have ⇒x=−2nπ+π2 \Rightarrow x = - 2n\pi + \dfrac{\pi }{2}
Here we have to find the solutions in the interval [0,2π)[0,2\pi ). Therefore x=3π2x = \dfrac{{3\pi }}{2}. Therefore, the solution or the equation sin⁡2x+sin⁡x=0{\sin ^2}x + \sin x = 0 in the interval [0,2π)[0,2\pi ) is x=0,3π2,πx = 0,\dfrac{{3\pi }}{2},\pi

Note: The trigonometry ratios have a value for the standard angles. The sine, cosine, tangent, cosecant, secant and cotangent are the trigonometry ratios of the trigonometry. The standard angles are 0, 30, 45, 60, 90, 180. By using these values we can determine the required solution.