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Question: How do you solve \(2{{\cos }^{2}}x+\sin x-2=0\) over the interval \(0\) to \(2\pi \)....

How do you solve 2cos⁡2x+sin⁡x−2=02{{\cos }^{2}}x+\sin x-2=0 over the interval 00 to 2π2\pi .

Explanation

Solution

In the given equation we have one cos⁡2{{\cos }^{2}} term and one sin⁡\sin term. So, we need to convert the whole equation in terms of cos⁡x\cos x or sin⁡x\sin x. For this we are going to use the trigonometric identity sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1. From this identity we will substitute the value of cos⁡2x{{\cos }^{2}}x as 1−sin⁡2x1-{{\sin }^{2}}x in the given equation and convert the given equation in form of ax2+bx+c=0a{{x}^{2}}+bx+c=0 . Now we will simplify and factorize the given equation, to get the solution of the given equation.

Complete step by step answer:
Given that, 2cos⁡2x+sin⁡x−2=02{{\cos }^{2}}x+\sin x-2=0.
We have the trigonometric identity sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1. From this identity the value of cos⁡2x{{\cos }^{2}}x is 1−sin⁡2x1-{{\sin }^{2}}x. Substituting this value in the given equation, then we will get
2(1−sin⁡2x)+sin⁡x−2=02\left( 1-{{\sin }^{2}}x \right)+\sin x-2=0
Simplifying the above equation, then we will get
2−2sin⁡2x+sin⁡x−2=0 ⇒−2sin⁡2x+sin⁡x=0 \begin{aligned} & 2-2{{\sin }^{2}}x+\sin x-2=0 \\\ & \Rightarrow -2{{\sin }^{2}}x+\sin x=0 \\\ \end{aligned}
Multiplying the above equation with negative sign, then we will get
2sin⁡2x−sin⁡x=02{{\sin }^{2}}x-\sin x=0
Taking sin⁡x\sin x common from the above equation, then we will get
⇒sin⁡x(2sin⁡x−1)=0\Rightarrow \sin x\left( 2\sin x-1 \right)=0
Equating each term to zero individually, then we will have
sin⁡x=0\sin x=0 or 2sin⁡x−1=0⇒sin⁡x=122\sin x-1=0\Rightarrow \sin x=\dfrac{1}{2}
In the problem they mentioned the interval as [0,2π]\left[ 0,2\pi \right].
So, in the given interval we will have sin⁡x=0\sin x=0 for x=\left\\{ 0,\pi ,2\pi \right\\}.
In the given interval we will have sin⁡x=12\sin x=\dfrac{1}{2} for x=\left\\{ \dfrac{\pi }{6},\dfrac{5\pi }{6} \right\\}

So, the set of solutions for the equation 2cos⁡2x+sin⁡x−2=02{{\cos }^{2}}x+\sin x-2=0 in the interval [0,2π]\left[ 0,2\pi \right] is \left\\{ 0,\dfrac{\pi }{6},\dfrac{5\pi }{6}\pi ,2\pi \right\\}.

Note: For this problem we can also convert sin⁡x\sin x into cos⁡2x{{\cos }^{2}}x and solve the problem. But there you will get a complex equation and the solution is not obtained easily and also there are a lot of chances to make mistakes. So, we have not chosen that method. You can also draw a graph for the given equation and check the points where the given equation meets the x−axisx-axis and those points are solutions for the equation.