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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  Answer Verified
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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