The value of x for the maximum value of (sqrt | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The expression is $f(x) = \sqrt{3} \sin x + \cos x$.We can rewrite this by multiplying and dividing by $\sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) The expression is $f(x) = \sqrt{3} \sin x + \cos x$.We can rewrite this by multiplying and dividing by $\sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2$.$f(x) = 2 \left( \frac{\sqrt{3}}{2} \sin x + \frac{1}{2} \cos x \right)$.Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we have $\sin(x + 30^\circ) = \sin x \cos 30^\circ + \cos x \sin 30^\circ = \frac{\sqrt{3}}{2} \sin x + \frac{1}{2} \cos x$.Thus,$f(x) = 2 \sin(x + 30^\circ)$.The maximum value of the sine function is $1$,which occurs when the angle is $90^\circ$.Therefore,$x + 30^\circ = 90^\circ$,which gives $x = 60^\circ$.

The value of $x$ for the maximum value of $(\sqrt{3} \sin x + \cos x)$ is .....$^o$

Similar Questions

If $f(x) = \cos^2 x + \sec^2 x$,then

If $\cot \theta + \tan \theta = m$ and $\sec \theta - \cos \theta = n,$ then which of the following is correct?

The value of $\cos \frac{\pi}{10} \cos \frac{2\pi}{10} \cos \frac{4\pi}{10} \cos \frac{8\pi}{10} \cos \frac{16\pi}{10}$ is:

The expression $\frac{\cos 6x + 6\cos 4x + 15\cos 2x + 10}{\cos 5x + 5\cos 3x + 10\cos x}$ is equal to

If $\theta$ is an acute angle and $\tan \theta + \cot \theta = 2$,then the value of $\tan^{5} \theta + \cot^{5} \theta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module