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Question: How do you solve: \(2{{\sin }^{2}}x-\sin x-1=0\)?...

How do you solve: 2sin⁡2x−sin⁡x−1=02{{\sin }^{2}}x-\sin x-1=0?

Explanation

Solution

The given equation is similar to a second-degree quadratic equation. Therefore we can factorize the equation to get the factors. And then by grouping the factors together, we can find out the root of this equation.

Complete Step by Step Solution:
The given trigonometric equation is
⇒2sin⁡2x−sin⁡x−1=0\Rightarrow 2{{\sin }^{2}}x-\sin x-1=0 ……(1)
Let us compare the given equation with the standard form of a second-degree quadratic equation.
⇒ax2+bx+c=0\Rightarrow a{{x}^{2}}+bx+c=0
As we can see, the given equation is matching up with the standard form of a second-degree quadratic equation. Hence we can factorize this like we usually do in a quadratic equation.
Equation (1) can be written as
⇒2sin⁡2x+sin⁡x−2sin⁡x−1=0\Rightarrow 2{{\sin }^{2}}x+\sin x-2\sin x-1=0
Because the component (sin⁡x−2sin⁡x)\left( \sin x-2\sin x \right) is equal to (−sin⁡x)\left( -\sin x \right)
Now let us take out the common variables
⇒sin⁡x(2sin⁡x+1)−(2sin⁡x+1)=0\Rightarrow \sin x\left( 2\sin x+1 \right)-\left( 2\sin x+1 \right)=0
If we take out (2sin⁡x+1)\left( 2\sin x+1 \right), we get
⇒(2sin⁡x+1)(sin⁡x−1)=0\Rightarrow \left( 2\sin x+1 \right)\left( \sin x-1 \right)=0 ……(2)
We have factored the equation (1) into two factors (2sin⁡x+1)\left( 2\sin x+1 \right) and (sin⁡x−1)\left( \sin x-1 \right).
When one of the factors becomes zero, equation (2) will be satisfied. So, let us equate each of the factors to zero, to find out the two possible roots of the given equation.
⇒2sin⁡x+1=0\Rightarrow 2\sin x+1=0
⇒sin⁡x=−12\Rightarrow \sin x=-\dfrac{1}{2}
Hence the variable x will be equal to
⇒x=sin⁡−1(−12)=7π6,11π6\Rightarrow x={{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{7\pi }{6},\dfrac{11\pi }{6}
The sine function is a periodic function with a time interval of 2π2\pi . Therefore
⇒x=7π6+2πn,11π6+2πn\Rightarrow x=\dfrac{7\pi }{6}+2\pi n,\dfrac{11\pi }{6}+2\pi n where n is an integer.
In a similar manner, find out the root of the factor (sin⁡x−1)\left( \sin x-1 \right), by equating it to zero.
sin⁡x−1=0\sin x-1=0
⇒sin⁡x=1\Rightarrow \sin x=1
⇒x=sin⁡−1(1)=π2\Rightarrow x={{\sin }^{-1}}\left( 1 \right)=\dfrac{\pi }{2}
Since the sine function is a periodic function with a time period of 2π2\pi ,
⇒x=π2+2πn\Rightarrow x=\dfrac{\pi }{2}+2\pi n, where n is an integer.

By combining both the roots, we can get the possible roots for the given trigonometric equation
⇒x=π2+2πn,7π6+2πn,11π6+2πn\Rightarrow x=\dfrac{\pi }{2}+2\pi n,\dfrac{7\pi }{6}+2\pi n,\dfrac{11\pi }{6}+2\pi n where n is an integer.

Note:
The trigonometric functions are periodic functions with a specific time interval for each function. While deducing the solution, we should always consider this periodic property of a trigonometric function and we should include it while calculating the answer.