baddinlol
Find the real values of x for which the equation is valid.
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baddinlol
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\[2\cos (x)=\sqrt{3}\cot (x)\]
Rosh007
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can you right cot (x) in terms of sin and cos
Rosh007
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***write nor right
baddinlol
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I wrote cot(x)=cos(x)/sin(x)
baddinlol
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And then i cancelled the cos(x)
Rosh007
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yup good
baddinlol
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And i was left with sin(x) = sqrt(3)/2
Rosh007
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now can you write what answer you got
baddinlol
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I then thought i can get infinite answers....
baddinlol
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So i wrote pi/3 and 2pi/3
baddinlol
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I think there are more solutions but i dont know how to writee them
Rosh007
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..... ok can you tell me what is th sin invers of \[\sin^{-1}( \frac{ \sqrt{3} }{ 2 })\]
baddinlol
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It is pi/3 and 2pi/3 between 0 and 2pi
Rosh007
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How you got 2pi/3
baddinlol
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I did pi - pi/3 because sin is also positive in that quadrant
Rosh007
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yes exactly.. so when we find inverse if get more values we take the lesser value usually.. for example 7pi/3 is also giving same answer but by using the concept above answer will b what?
baddinlol
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is it pi/3?
Rosh007
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yup it is
baddinlol
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But i need to state all the values for it
baddinlol
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so i need to write, pi/3, 2pi/3, 7pi/3, 8pi/3, 13pi/3, 14pi/3 .......
baddinlol
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How can i write this as a general solution
Rosh007
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ok.. first take a look at pi/3 and 7pi/3 what can you find
Rosh007
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any similarity?.. or relation... (clue can you relate with multiple of 2pi
baddinlol
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Oo i think i found it
baddinlol
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Is it pi/3 + 2kpi
baddinlol
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and 2pi/3 + 2kpi
Rosh007
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exactly.. ther you go... what is k...( no need to tll if you ar pretty sure)
baddinlol
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And one more thing
baddinlol
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Will i also need to evaluate cos(x)=0?
Rosh007
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??? not clear... cab you make it more clear?is it another question?
Rosh007
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** can
Rosh007
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if it is another qustion your answer is (2k+1)pi/2
baddinlol
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No same one
baddinlol
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wouldn't 2cos(x) = sqrt(3)cot(x) if cos(x) = 0 also
Rosh007
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am I clear?
Rosh007
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ok...... asuming a fuction itself is zero may not lad you to corrct ansewr... also... here the cos (x) cancels each other (ofcourse before assuming it zero) hence assuming cos (x) = 0 is wrong..
Rosh007
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**lead you to correct answer