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$\cos 2\theta + 2\cos \theta $ is always?
$A{\text{.}}$ Greater than $\frac{{ - 3}}{2}$
$B.$ Less than or equal to $\frac{3}{2}$
${\text{C}}{\text{.}}$ Greater than or equal to $\frac{{ - 3}}{2}$ and less than or equal to 3
${\text{D}}{\text{.}}$ None of the above

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Hint:-Here, we go through the formula $\cos 2\theta = 2{\cos ^2}\theta - 1$ we have to make $2\theta $ in terms of $\theta $. And then by applying its range we get the extreme values or minimum values.

Given,
$\cos 2\theta + 2\cos \theta $
$ \Rightarrow \cos 2\theta + 2\cos \theta = 2{\cos ^2}\theta - 1 + 2\cos \theta $
Here we use the formula $\cos 2\theta = 2{\cos ^2}\theta - 1$
We can rewrite above question as following:
$ \Rightarrow \cos 2\theta + 2\cos \theta = 2{\left( {\cos \theta + \frac{1}{2}} \right)^2} - \frac{3}{2}$
Now, we know that the square of every number is greater than or equal to zero.
So, ${\text{2}}{\left( {\cos \theta + \frac{1}{2}} \right)^2} \geqslant 0$ for all $\theta $.
$\therefore 2{\left( {\cos \theta + \frac{1}{2}} \right)^2} - \frac{3}{2} \geqslant \frac{{ - 3}}{2}$ For all $\theta $.
$ \Rightarrow {\text{cos2}}\theta {\text{ + 2cos}}\theta \geqslant \frac{{ - 3}}{2}$ For all $\theta $.
And we can clearly see maximum value is $3$ when $\theta {\text{ = }}{{\text{0}}^0}$
Hence, option C is the correct answer.

Note: - when you got questions regarding maximum or minimum value. Then you have to proceed after making a square of terms and then after checking at extreme values.

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