Find the maximum value of trigonometry function

Question

Find the maximum value of
$$\frac{1}{1+\frac{3\sin2\theta}{2}+4(cos\theta)^2}$$

Answer 1

We need minimum positive value of $$1+\dfrac{3\sin2\theta}2+4\cos^2\theta$$

$$=\dfrac{2+3\sin2\theta+4(1+\cos2\theta)}2$$

Now $3\sin2\theta+4\cos2\theta\ge-\sqrt{3^2+4^2}$

Answer 2

HINT:

$1)$Differentiate the function one time.

$2)$then by equaling the derivative to zero you will get the critical points.

$3)$then put them in the second derivative one by one. For which the number will be more negative is the maximum one.

$4)$Then use that value of "$\theta$" to get the maximum value.

Answer 3

Hint:

Let $y=\dfrac{1}{1+\dfrac{3\sin2\theta}{2}+4(\cos\theta)^2}$

$$y+3y\sin\theta\cos\theta+4y\cos^2\theta=1$$

Divide both sides by $\cos^2\theta,$

$$3y\tan\theta+4y=(1-y)(1+\tan^2\theta)$$ which is a quadratic equation $\tan\theta$

As the $\tan\theta$ is real, the discriminant mist be $\ge0$