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[Solved] Range of the trigonometric function

 
Trigonometry 1 & 2
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June 15, 2021 3:11 pm
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(@armor)
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June 15, 2021 3:12 pm
Topic starter
Staff 
(@armor)
Admin
  Honorable Member
317 Posts
156 153 0

According to the domain of square root, we can obtain \small 0\leq x^2+x+1\leq1.   

Since \small x^2+x+1\leq 1,  then \small -1\leq x\leq 0.  Hence, the domain of the function is \small [-1,0].

Since \small x^2+x+1=(x+\frac{1}{2})^2+\frac{3}{4}\geq \frac{3}{4}\Rightarrow \frac{3}{4}\leq x^2+x+1\leq 1

\small \Rightarrow 0\leq arccos(x^2+x+1)\leq arccos\frac{3}{4}\Rightarrow 0\leq \sqrt{arccos(x^2+x+1)}\leq \sqrt{arccos\frac{3}{4}}.

Therefore, the range of the function is \small [0, \sqrt{arccos\frac{3}{4}}].

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