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[Solved] Trigonometric functions

 
Trigonometry 1 & 2
Last Post by Staff 4 years ago
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June 15, 2021 3:03 pm
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(@armor)
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June 15, 2021 3:04 pm
Topic starter
Staff 
(@armor)
Admin
  Honorable Member
317 Posts
156 153 0

\small f(0)=2a=2\Rightarrow a=1.  Since     \small f(\frac{\Pi }{3})=\frac{1}{2}a+\frac{\sqrt{3}}{4}b=\frac{1}{2}+\frac{\sqrt{3}}{2},

substituting \small a=1 into this formula, we have \small b=2.  Thus

\small f(x)=2cos^2x+2sinxcosx=sin2x+cos2x+1.

Since \small f(x)> 2,  then  \small sin2x+cos2x+1>2\Rightarrow sin(2x+\frac{\Pi }{4})> \frac{\sqrt{2}}{2}

\small \Rightarrow 2k\Pi +\frac{\Pi }{4}<(2x+\frac{\Pi }{4})<2k\Pi +\frac{3\Pi }{4} (k\epsilon \mathbb{Z}).

Therefore the set of \small x values that satisfy the formula \small f(x)>2 is 

\small (x|k\Pi <x<k\Pi +\frac{\Pi }{4}, k\epsilon \mathbb{Z})

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