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Question: Solve the following equation \[2{{\sin }^{2}}\theta =3\cos \theta \], \[0\le \theta \le 2\pi \]....

Solve the following equation 2sin2θ=3cosθ2{{\sin }^{2}}\theta =3\cos \theta , 0θ2π0\le \theta \le 2\pi .

Explanation

Solution

Hint: In the above equation we will use the trigonometric identity which given as follows:
sin2θ+cos2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1
After using the identity, we will solve the quadratic equation in cosθ\cos \theta and we will get the solution for the equation.

Complete Step-by-step answer:
We have been given the equation is 2sin2θ=3cosθ2{{\sin }^{2}}\theta =3\cos \theta .
As we know that sin2θ=1cos2θ{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta .
So by using this in the above equation, we get as follows:

& 2\left( 1-{{\cos }^{2}}\theta \right)=3\cos \theta \\\ & 2-2{{\cos }^{2}}\theta =3\cos \theta \\\ \end{aligned}$$ Take all the terms of left hand side of the equation to right hand side, we get as follows: $$\begin{aligned} & 0=3\cos \theta +2{{\cos }^{2}}\theta -2 \\\ & 2{{\cos }^{2}}\theta +3\cos \theta -2 \\\ \end{aligned}$$ Now we can write $$3\cos \theta $$ in the form of $$\left( 4\cos \theta -\cos \theta \right)$$ in order to factorize the above quadratic equation in ‘$$\cos \theta $$’. $$\begin{aligned} & 2{{\cos }^{2}}\theta +4\cos \theta -\cos \theta -2=0 \\\ & 2\cos \theta \left( \cos \theta +2 \right)-1\left( \cos \theta +2 \right)=0 \\\ \end{aligned}$$ After taking $$\left( \cos \theta +2 \right)$$ as common from the above equation, we get as follows: $$\left( \cos \theta +2 \right)\left( 2\cos \theta -1 \right)=0$$ So $$\Rightarrow \cos \theta +2=0$$ and $$2\cos \theta -1=0$$. As we know the range of $$\cos \theta $$ is $$\left[ -1,1 \right]$$. $$\Rightarrow \cos \theta \ne -2$$ So we have only $$2\cos \theta -1=0$$. $$\Rightarrow \cos \theta =\dfrac{1}{2}$$ As we know that the general solution of $$\cos \theta =\cos \alpha $$ is given by $$\theta =2n\pi \pm \alpha $$ where ‘n’ is any integer. $$\begin{aligned} & \cos \theta =\cos \dfrac{\pi }{3} \\\ & \theta =2n\pi \pm \dfrac{\pi }{3} \\\ \end{aligned}$$ As we have been given that $$0\le \theta \le 2\pi $$. $$\Rightarrow \theta =\dfrac{\pi }{3},\dfrac{5\pi }{3}$$ Therefore, the solution for the given equations are $$\theta =\dfrac{\pi }{3}$$ and $$\theta =\dfrac{5\pi }{3}$$. Note: Be careful while finding the exact solution of the equation as we have given that $$0\le \theta \le 2\pi $$. Also, be careful while solving the quadratic equation in $$\cos \theta $$ as there is a chance that you might make a mistake while taking the common terms out. We can also find the exact solution by drawing the graph of $$\cos \theta =\dfrac{1}{2}$$ for $$0\le \theta \le 2\pi $$.