Question

3) = x + 2{\rm{sin}}x\)

Short answer

The required value is \(x = \left( {2n + 1} \right)\pi \pm \frac{\pi }{3},\,\,\,\,\,\,\,{\rm{for}}\,\,n \in {\rm I}\).

Step-by-step solution

 Step 1: Horizontal Tangents of Trigonometric Functions

When any trigonometric function is subjected to have horizontal tangents, then thederivativeof that trigonometric function will be equal to zero.

Differentiation ofthe given function and solving for \(x\) 

The given function is\(f\left( x \right) = x + 2\sin x\).

Differentiating the given functionwith respect to\(x\)and equating it to zero, we get:

\(\begin{array}{c}\frac{d}{{dx}}\left( {f\left( x \right)} \right) = 0\\\frac{d}{{dx}}\left( {x + 2\sin x} \right) = 0\\1 + 2\cos x = 0\\\cos x = - \frac{1}{2}\\x = \left( {\frac{{2\pi }}{3} + 2\pi n} \right)\,\,{\rm{or }}\left( {\frac{{4\pi }}{3} + 2\pi n} \right)\,\,\,\,\,\,\forall n \in {\rm I}\\x = \left( {2n + 1} \right)\pi \pm \frac{\pi }{3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\forall n \in {\rm I}\end{array}\)

Hence, the required answer is\(x = \left( {2n + 1} \right)\pi \pm \frac{\pi }{3},\,\,\,\,\,\,\,{\rm{for}}\,\,n \in {\rm I}\).