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Question: How do you solve the equation \(2{{\cos }^{2}}\theta -{{\sin }^{2}}\theta =1\) ?...

How do you solve the equation 2cos⁡2θ−sin⁡2θ=12{{\cos }^{2}}\theta -{{\sin }^{2}}\theta =1 ?

Explanation

Solution

We have been given an equation consisting of two trigonometric functions, sine function and cosine function. Firstly, we shall convert the entire equation in terms of the sine function using the trigonometric identities. Then we shall obtain a quadratic equation which will be solved further like a quadratic function by substituting it equal to some variable-t.

Complete step by step solution:
Given that 2cos⁡2θ−sin⁡2θ=12{{\cos }^{2}}\theta -{{\sin }^{2}}\theta =1
We shall use the trigonometric identity sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1 to substitute the square of cosine function with 1 minus the square of sine function, that is, cos⁡2x=1−sin⁡2x{{\cos }^{2}}x=1-{{\sin }^{2}}x.
⇒2(1−sin⁡2θ)−sin⁡2θ=1\Rightarrow 2\left( 1-{{\sin }^{2}}\theta \right)-{{\sin }^{2}}\theta =1
⇒2−2sin⁡2θ−sin⁡2θ=1 ⇒2−3sin⁡2θ=1 \begin{aligned} & \Rightarrow 2-2{{\sin }^{2}}\theta -{{\sin }^{2}}\theta =1 \\\ & \Rightarrow 2-3{{\sin }^{2}}\theta =1 \\\ \end{aligned}
Transposing the constant term 1 to the left hand side of equation, we get
⇒2−3sin⁡2θ−1=0\Rightarrow 2-3{{\sin }^{2}}\theta -1=0
⇒1−3sin⁡2θ=0\Rightarrow 1-3{{\sin }^{2}}\theta =0
We will multiply the entire equation with a negative sign.
⇒3sin⁡2θ−1=0\Rightarrow 3{{\sin }^{2}}\theta -1=0
Now, we shall substitute sine function equal to some variable-t.
Let t=sin⁡θt=\sin \theta
⇒3t2−1=0\Rightarrow 3{{t}^{2}}-1=0
Here, we will transpose the constant term 1 to the right hand side and then divide the entire equation by 3 to make the coefficient of variable-t equal to 1.
⇒3t2=1\Rightarrow 3{{t}^{2}}=1
⇒t2=13\Rightarrow {{t}^{2}}=\dfrac{1}{3}
Further, square rooting the terms on both sides of the equation, we get
⇒t2=±13\Rightarrow \sqrt{{{t}^{2}}}=\pm \sqrt{\dfrac{1}{3}}
⇒t=±13\Rightarrow t=\pm \dfrac{1}{\sqrt{3}}
Now, resubstituting the value t=sin⁡θt=\sin \theta , we get
⇒sin⁡θ=±13\Rightarrow \sin \theta =\pm \dfrac{1}{\sqrt{3}}
⇒sin⁡θ=13\Rightarrow \sin \theta =\dfrac{1}{\sqrt{3}} and sin⁡θ=−13\sin \theta =-\dfrac{1}{\sqrt{3}}.
The values of θ\theta for which the above equation is satisfied is 0.615 and -0.615 respectively.
Therefore, the general solution of 2cos⁡2θ−sin⁡2θ=12{{\cos }^{2}}\theta -{{\sin }^{2}}\theta =1 is θ∈nπ+(−1)n0.615\theta \in n\pi +{{\left( -1 \right)}^{n}}0.615.

Note: In order to find the solution of various trigonometric equations, we must have prior knowledge of the main trigonometric identities. Another method of solving this problem was by drawing the graphs of sine function and the horizontal lines y=13y=\dfrac{1}{\sqrt{3}} and y=−13y=-\dfrac{1}{\sqrt{3}} on the same graph and then finding the points of intersection of sine graph with these lines.