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haleyelizabeth2017
 one year ago
What values for θ (0≤θ≤2π) satisfy the equation?
haleyelizabeth2017
 one year ago
What values for θ (0≤θ≤2π) satisfy the equation?

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haleyelizabeth2017
 one year ago
Best ResponseYou've already chosen the best response.0\[tan^2\theta=\frac{3}{2}sec\theta\]

acxbox22
 one year ago
Best ResponseYou've already chosen the best response.0sorry i suck at trig :(

haleyelizabeth2017
 one year ago
Best ResponseYou've already chosen the best response.0It's all good :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do U know the identity that relates both of the \[\sin (\Theta) , \tan (\Theta)\] ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sec ^2 \theta \tan ^2 \theta=1\]

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1this 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
 one year ago
Best ResponseYou've already chosen the best response.0\[\tan ^2 \theta=\sec ^2 \theta1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sec ^2 \theta 1=\frac{ 3 }{ 2 }\cos \theta\] \[2 \sec ^2 \theta2=3 \sec \theta\] solve for sec theta

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1when 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

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1oops should be will give 2 exact values... this will allow you to find the angles... hope it helps

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So , 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
 one year ago
Best ResponseYou've already chosen the best response.0Ah! sorry...was afk :(

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sec ^2 \theta1=\frac{ 3 }{ 2 }\sec \theta\] \[2\sec ^2\theta+3\sec \theta2=0\] \[2\sec ^2\theta+4\sec \theta\sec \theta2=0\] \[2\sec \theta \left( \sec \theta+2 \right)1\left( \sec \theta+2 \right)=0\] \[\left( \sec \theta+2 \right)\left( 2\sec \theta1 \right)=0\] \[Either~\sec \theta+2=0,\sec \theta=2,\cos \theta=\frac{ 1 }{ 2 }=\cos \frac{ \pi }{ 3 }=\cos \left( \pi \pm \frac{ \pi }{ 3 } \right)\] \[\theta=\pi \pm \frac{ \pi }{ 3 }\] \[or~2\sec \theta1=0.\sec \theta=\frac{ 1 }{ 2 },\cos \theta=2\] rejected as\[\left \cos \theta \right\le 1\]
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