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Question: How do you solve \(2{{\sin }^{2}}x\ge \sin x\) using a sign chart?...

How do you solve 2sin⁡2x≥sin⁡x2{{\sin }^{2}}x\ge \sin x using a sign chart?

Explanation

Solution

We have been given an inequality equation consisting of the trigonometric function sine of x. In order to find the solution of this equation, we must compute the values of x for which this inequality holds true. We shall first simplify this equation by grouping and find the values for which it is equal to zero. Then, we will mark these intervals to make the sign chart and find our solution.

Complete step by step solution:
Given that, 2sin⁡2x≥sin⁡x2{{\sin }^{2}}x\ge \sin x
Transposing sinx to the left hand side, we get
⇒2sin⁡2x−sin⁡x≥0\Rightarrow 2{{\sin }^{2}}x-\sin x\ge 0
Here, we shall take sinx common and then group the remaining terms together.
⇒sin⁡x(2sin⁡x−1)≥0\Rightarrow \sin x\left( 2\sin x-1 \right)\ge 0
Separately equating these two groups equal to zero, we get
2sin⁡x−1=02\sin x-1=0 and sin⁡x=0\sin x=0
⇒sin⁡x=12\Rightarrow \sin x=\dfrac{1}{2} and sin⁡x=0\sin x=0
For sin⁡x=12\sin x=\dfrac{1}{2},
x=π6,5π6x=\dfrac{\pi }{6},\dfrac{5\pi }{6}
For sin⁡x=0\sin x=0,
x=0,π,2πx=0,\pi ,2\pi
Now, we shall make the sign chart of the given equation f(x)=2sin⁡2x−sin⁡xf\left( x \right)=2{{\sin }^{2}}x-\sin x.

xx(0,π6)\left( 0,\dfrac{\pi }{6} \right)(π6,5π6)\left( \dfrac{\pi }{6},\dfrac{5\pi }{6} \right)(5π6,π)\left( \dfrac{5\pi }{6},\pi \right)(π,2π)\left( \pi ,2\pi \right)
2sin⁡x−12\sin x-1-+--
sin⁡x\sin x+++-
f(x)f\left( x \right)-+-+

Thus, x∈[π6,5π6]∪[π,2π]x\in \left[ \dfrac{\pi }{6},\dfrac{5\pi }{6} \right]\cup \left[ \pi ,2\pi \right] for f(x)≥0f\left( x \right)\ge 0, that is, for 2sin⁡2x−sin⁡x≥02{{\sin }^{2}}x-\sin x\ge 0.

Therefore, the general solution of given equation is x∈[π6+2nπ,5π6+2nπ]∪[π+2nπ,2π+2nπ]x\in \left[ \dfrac{\pi }{6}+2n\pi ,\dfrac{5\pi }{6}+2n\pi \right]\cup \left[ \pi +2n\pi ,2\pi +2n\pi \right].

Note: The easiest way to determine where a function is positive or negative is by using a sign chart. A function is positive for values for which it lies above the x-axis and a function is negative for values for which it lies below the x-axis. A function has to cross the x-axis once as it changes its sign from positive to negative or vice versa. Thus, we find the value of our function for which it is equal to zero. After obtaining the values for which our given function is equal to zero, we construct the skeleton of a number line and mark these values on the number line and the intervals on the number line accordingly. Further, we check the sign of the function in that interval and mark that interval as positive or negative.