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Question: How do you solve \[2{{\sin }^{2}}\left( a \right)=2+\cos \left( a \right)\] and find all solutions i...

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

Explanation

Solution

In order to find the solution to the given question that is to solve 2sin⁡2(a)=2+cos⁡(a)2{{\sin }^{2}}\left( a \right)=2+\cos \left( a \right) and find all solutions in the interval [0,2π)\left[ 0,2\pi \right), apply the one of the well-known identities of trigonometry that is sin⁡2(a)=1−cos⁡2(a){{\sin }^{2}}\left( a \right)=1-{{\cos }^{2}}\left( a \right). Substitute this value in the given expression and then find the value of cos⁡(a)\cos \left( a \right) followed by the finding the value of aa in the interval [0,2π)\left[ 0,2\pi \right).

Complete step-by-step solution:
According to the question, given equation in the question is as follows:
2sin⁡2(a)=2+cos⁡(a)2{{\sin }^{2}}\left( a \right)=2+\cos \left( a \right)
To solve the above equation, apply one of the well-known identities of trigonometry that is sin⁡2(a)=1−cos⁡2(a){{\sin }^{2}}\left( a \right)=1-{{\cos }^{2}}\left( a \right) & substitute in the above equation, we get:
⇒2(1−cos⁡2(a))=2+cos⁡(a)\Rightarrow 2\left( 1-{{\cos }^{2}}\left( a \right) \right)=2+\cos \left( a \right)
Simplify the terms in the left-hand side of the above equation by opening the bracket we get:
⇒2−2cos⁡2(a)=2+cos⁡(a)\Rightarrow 2-2{{\cos }^{2}}\left( a \right)=2+\cos \left( a \right)
As we can see term 22 can be cancelled from both the sides in the above equation, we get:
⇒−2cos⁡2(a)−cos⁡(a)=0\Rightarrow -2{{\cos }^{2}}\left( a \right)-\cos \left( a \right)=0
Now, take the term −cos⁡(a)-\cos \left( a \right)in common and rest the terms in the bracket from the above equation, we will have:
⇒−cos⁡(a)(2cos⁡(a)+1)=0\Rightarrow -\cos \left( a \right)\left( 2\cos \left( a \right)+1 \right)=0
From above equation we can infer two results that are:

  1. cos⁡(a)=0\cos \left( a \right)=0
    ⇒a=π2\Rightarrow a=\dfrac{\pi }{2} and a=3π2a=\dfrac{3\pi }{2}
  2. 2cos⁡(a)+1=02\cos \left( a \right)+1=0
    ⇒cos⁡(a)=−12\Rightarrow \cos \left( a \right)=-\dfrac{1}{2}
    ⇒a=±2π3\Rightarrow a=\pm \dfrac{2\pi }{3}
    The arc 2π3\dfrac{2\pi }{3} and arc 4π3\dfrac{4\pi }{3} are coterminal.
    Therefore, the values of aa in the interval [0,2π)\left[ 0,2\pi \right) are π2,3π2,2π3 !!&!! 4π3\dfrac{\pi }{2},\dfrac{3\pi }{2},\dfrac{2\pi }{3}\text{ }\\!\\!\And\\!\\!\text{ }\dfrac{4\pi }{3}.

Note: Students generally make mistakes while finding the value of angle of cosine or any trigonometric function by just finding out one value, but it’s actually two answers. It’s important to check for the 2nd value and don’t just ignore it because it gives precision to the result of your answer and it’s important the remember the identity sin⁡2(a)=1−cos⁡2(a){{\sin }^{2}}\left( a \right)=1-{{\cos }^{2}}\left( a \right).