haleyelizabeth2017
  • haleyelizabeth2017
What values for θ (0≤θ≤2π) satisfy the equation?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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haleyelizabeth2017
  • haleyelizabeth2017
\[tan^2\theta=-\frac{3}{2}sec\theta\]
acxbox22
  • acxbox22
sorry i suck at trig :(
haleyelizabeth2017
  • haleyelizabeth2017
It's all good :)

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anonymous
  • anonymous
do U know the identity that relates both of the \[\sin (\Theta) , \tan (\Theta)\] ?
anonymous
  • anonymous
\[\sec ^2 \theta -\tan ^2 \theta=1\]
campbell_st
  • campbell_st
this is how to start \[\tan^2(\theta) = \frac{\sin^2(\theta)}{\cos^2(\theta)}\] and \[\sec(\theta) = \frac{1}{\cos(\theta)}\] multiply both sides of the equation by cos^2 \[\sin^2(\theta) = \frac{3}{2} \cos(\theta)\] now use the trig identity sin^2 + cos^2 = 1 so \[1 - \cos^2(\theta) = \frac{3}{2} \cos(\theta)\] which can be written as \[\cos^2(\theta) + \frac{3}{2}\cos(\theta) - 1 = 0\] so now you have a quadratic that can be solved... hope it helps
anonymous
  • anonymous
\[\tan ^2 \theta=\sec ^2 \theta-1\]
anonymous
  • anonymous
\[\sec ^2 \theta -1=\frac{ 3 }{ 2 }\cos \theta\] \[2 \sec ^2 \theta-2=3 \sec \theta\] solve for sec theta
campbell_st
  • campbell_st
when you solve the quadratic using the general quadratic formula you are solving \[\cos(\theta) = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}\] this will give an exact value, then find the angle measures
anonymous
  • anonymous
$\theta$
anonymous
  • anonymous
$tan^2$
campbell_st
  • campbell_st
oops should be will give 2 exact values... this will allow you to find the angles... hope it helps
anonymous
  • anonymous
So , I'll assume you don't know it :D \[\tan^2 (\Theta) +1 = \sec^2 (\Theta)\] so nw U have \[\tan^2 (\Theta) = \sec^2 (\Theta) -1\] Simply you plug this into you equation and you will get sth like this : \[\sec^2 (\Theta) -1 = -\frac{ 3 }{ 2 } \sec(\Theta)\] You can easily get the \[-\frac{ 3 }{ 2 } \sec(\Theta)\] to the right hand side by changing the sign :) The you will get : \[\sec^2(\Theta) + 1.5 \sec(\Theta) - 1 = 0\] nw guess what ! It's a quadratic equation that could be solved easily Ur calculator or the formula and the you will get \[\sec(\Theta) = \frac{ 1 }{ 2 } , -2\] This is for sec to get theta you have to flip the answer and you will get \[\cos(\Theta) = -\frac{ 1 }{ 2 } , 2\] So the second one will be rejected as 2 is not a value for any Theta ! nw you will get Theta simply like this : \[\Theta = \cos^{-1} (-\frac{ 1 }{ 2 }) = 120^o = \frac{ 2 }{ 3 }π\] radians :D Let me know if you got it :)
haleyelizabeth2017
  • haleyelizabeth2017
Ah! sorry...was afk :(
anonymous
  • anonymous
\[\sec ^2 \theta-1=-\frac{ 3 }{ 2 }\sec \theta\] \[2\sec ^2\theta+3\sec \theta-2=0\] \[2\sec ^2\theta+4\sec \theta-\sec \theta-2=0\] \[2\sec \theta \left( \sec \theta+2 \right)-1\left( \sec \theta+2 \right)=0\] \[\left( \sec \theta+2 \right)\left( 2\sec \theta-1 \right)=0\] \[Either~\sec \theta+2=0,\sec \theta=-2,\cos \theta=-\frac{ 1 }{ 2 }=-\cos \frac{ \pi }{ 3 }=\cos \left( \pi \pm \frac{ \pi }{ 3 } \right)\] \[\theta=\pi \pm \frac{ \pi }{ 3 }\] \[or~2\sec \theta-1=0.\sec \theta=\frac{ 1 }{ 2 },\cos \theta=2\] rejected as\[\left| \cos \theta \right|\le 1\]

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