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  • What is the range of x  for <img alt="\sin2x >|\cos x| \ where\ x\epsilon[-\frac{\pi}{2},\frac{\pi}{2}]" src="https://learn.careers360.com/latex-image/?%5Csin2x%20%3E%7C%5Ccos%20x%7C%2

What is the range of x  for \sin2x >|\cos x| \ where\ x\epsilon[-\frac{\pi}{2},\frac{\pi}{2}]

Option: 1

x\epsilon [\frac{\pi}{6},\frac{\pi}{2}]


Option: 2

x\epsilon [\frac{\pi}{3},\frac{\pi}{2}]


Option: 3

x\epsilon [-\frac{\pi}{2},-\frac{\pi}{6}]U[\frac{\pi}{2},\frac{\pi}{6}]


Option: 4

None of these 


Answers (1)

best_answer

Trigonometric Inequality -

Trigonometric Inequality

 

The trigonometric inequation of the type f(x) ≥ a or f(x) ≤ a, where f(x) is some trigonometric ratio, the following steps should be taken to solve such type of equations.:

  1. Draw the graph of f(x) in an interval length equal to the fundamental period of f(x) .

  2. Draw the line y = a .

  3. Take the portion of the graph for which the inequation is satisfied.

  4. To generalize, add nT (n ∈ I) and take union over the set of integers, where T is fundamental period of f(x).

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\sin2x >|\cos x| \ where\ x\epsilon[-\frac{\pi}{2},\frac{\pi}{2}]

For the intersection point of sin2x and cos x

\sin 2x=\cos x\\ 2\sin x \cos x=\cos x\\ \cos x(2 \sin x- 1)=0\\ \cos x=0 \rightarrow x=\frac{\pi}{2}\\ \sin x =\frac{\pi}{6}

By the Graph, We can see that this inequality satisfy when x\epsilon [\frac{\pi}{6},\frac{\pi}{2}]

Posted by

Sanket Gandhi

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