Question

The general solution of the equation, $2\cos 2x=3.2{{\cos }^{2}}x-4$ .
A. $2n\pi ,n\in I$
B. $n\pi ,n\in I$
C. $\dfrac{n\pi }{4},n\in I$
D. $\dfrac{n\pi }{2},n\in I$

Answer 1

Hint: In the question, we have to find the general solution of $2\cos 2x=3.2{{\cos }^{2}}x-4$ . We know the formula that, $\cos 2x=2{{\cos }^{2}}x-1$ . Using this formula, replace $2{{\cos }^{2}}x$ by $\cos 2x+1$ . Now, find the value of $\cos 2x$ . We get $\cos 2x=1$ . We can write it as, $\cos 2x=\cos 2n\pi$ .The general solution of $\cos 2x=1$ is $x=n\pi$ .

Complete step-by-step answer:
According to the question, we have an equation,
$2\cos 2x=3.2{{\cos }^{2}}x-4$ …………….(1)
We know the formula, $\cos 2x=2{{\cos }^{2}}x-1$ …………………(2)
On solving equation (2), we get
$\cos 2x=2{{\cos }^{2}}x-1$
$\Rightarrow \cos 2x-1=2{{\cos }^{2}}x$ …………………..(3)
From equation (1) and equation (3), we get
$2\cos 2x=3.2{{\cos }^{2}}x-4$
$\Rightarrow 2\cos 2x=3(1+\cos 2x)-4$
Solving it further, we get

& \Rightarrow 2\cos 2x=3+3\cos 2x-4 \\\ & \Rightarrow 2\cos 2x=3\cos 2x-1 \\\ & \Rightarrow 1=3\cos 2x-2\cos 2x \\\ \end{aligned}$$ $$\Rightarrow 1=\cos 2x$$ …………………(4) We know that $$\cos 0=1$$ . Now using this in equation (4), we get $$\begin{aligned} & \cos 2x=1 \\\ & \Rightarrow \cos 2x=\cos 0 \\\ & \Rightarrow 2x=0 \\\ \end{aligned}$$ We also know that $$\cos 2\pi =1$$ . Now using this in equation (4), we get $$\begin{aligned} & \cos 2x=1 \\\ & \Rightarrow \cos 2x=\cos 2\pi \\\ & \Rightarrow 2x=2\pi \\\ \end{aligned}$$ We also know that, $$\cos 4\pi =1$$ . Now using this in equation (4), we get $$\begin{aligned} & \cos 2x=1 \\\ & \Rightarrow \cos 2x=\cos 4\pi \\\ & \Rightarrow 2x=4\pi \\\ \end{aligned}$$ We can see that x can be 0, $$2\pi $$ , $$4\pi $$ , $$6\pi $$ etc. We can observe that x is the multiple of $$2\pi $$ ……………….(5) Now, from equation (4) and equation (5), we get $$\cos 2x=1$$ $$\begin{aligned} & \Rightarrow \cos 2x=\cos 2n\pi \\\ & \Rightarrow 2x=2n\pi \\\ & \Rightarrow x=n\pi \\\ \end{aligned}$$ Hence, option (B) is the correct one. Note: We can also solve this question by using another method. We know the formula, $$\cos 2x=2{{\cos }^{2}}x-1$$ . Now putting this in $$2\cos 2x=3.2{{\cos }^{2}}x-4$$ . $$\begin{aligned} & 2\cos 2x=3.2{{\cos }^{2}}x-4 \\\ & \Rightarrow 2(2{{\cos }^{2}}x-1)=6{{\cos }^{2}}x-4 \\\ & \Rightarrow 4{{\cos }^{2}}x-2=6{{\cos }^{2}}x-4 \\\ & \Rightarrow 2({{\cos }^{2}}x-1)=0 \\\ & \Rightarrow (\cos x+1)(\cos x-1)=0 \\\ \end{aligned}$$ $$\cos x=-1$$ ……………(1) or $$\cos x=1$$ ……………………(2) We know that $$\cos 0=1$$ . Now using this in equation (1), we get $$\begin{aligned} & \cos x=1 \\\ & \Rightarrow \cos x=\cos 0 \\\ & \Rightarrow x=0 \\\ \end{aligned}$$ We also know that $$\cos 2\pi =1$$ . Now using this in equation (1), we get $$\begin{aligned} & \cos x=1 \\\ & \Rightarrow \cos x=\cos 2\pi \\\ & \Rightarrow x=2\pi \\\ \end{aligned}$$ We also know that, $$\cos 4\pi =1$$ . Now using this in equation (1), we get $$\begin{aligned} & \cos x=1 \\\ & \Rightarrow \cos x=\cos 4\pi \\\ & \Rightarrow x=4\pi \\\ \end{aligned}$$ We can see that x can be 0, $$2\pi $$ , $$4\pi $$ , $$6\pi $$ etc.……………………(3) If $$\cos x=1$$ , then x is the multiple of $$2\pi $$ . We know that $$\cos \pi =-1$$ . Now using this in equation (2), we get $$\begin{aligned} & \cos x=-1 \\\ & \Rightarrow \cos x=\cos \pi \\\ & \Rightarrow x=\pi \\\ \end{aligned}$$ We also know that $$\cos 3\pi =-1$$ . Now using this in equation (2), we get $$\begin{aligned} & \cos x=-1 \\\ & \Rightarrow \cos x=\cos 3\pi \\\ & \Rightarrow x=3\pi \\\ \end{aligned}$$ We also know that, $$\cos 5\pi =-1$$ . Now using this in equation (2), we get $$\begin{aligned} & \cos x=-1 \\\ & \Rightarrow \cos x=\cos 5\pi \\\ & \Rightarrow x=5\pi \\\ \end{aligned}$$ We can see that x can be $$\pi $$ , $$3\pi $$ , $$5\pi $$ etc. ……………….(4) We can observe that x is the odd multiple of $$\pi $$. If $$\cos x=-1$$ , then x is the odd multiple of $$\pi $$. From equation (3) and (4), we have 0, $$\pi $$ , $$2\pi $$ , $$3\pi $$ , $$4\pi $$ , $$5\pi $$ , and $$6\pi $$ . Therefore, x is the multiple of $$\pi $$ . Hence, option (B) is the correct one.