What values for θ (0≤θ≤2π) satisfy the equation?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

What values for θ (0≤θ≤2π) satisfy the equation?

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\[tan^2\theta=-\frac{3}{2}sec\theta\]
sorry i suck at trig :(
It's all good :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

do U know the identity that relates both of the \[\sin (\Theta) , \tan (\Theta)\] ?
\[\sec ^2 \theta -\tan ^2 \theta=1\]
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
\[\tan ^2 \theta=\sec ^2 \theta-1\]
\[\sec ^2 \theta -1=\frac{ 3 }{ 2 }\cos \theta\] \[2 \sec ^2 \theta-2=3 \sec \theta\] solve for sec theta
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
$\theta$
$tan^2$
oops should be will give 2 exact values... this will allow you to find the angles... hope it helps
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 :)
Ah! sorry...was afk :(
\[\sec ^2 \theta-1=-\frac{ 3 }{ 2 }\sec \theta\] \[2\sec ^2\theta+3\sec \theta-2=0\] \[2\sec ^2\theta+4\sec \theta-\sec \theta-2=0\] \[2\sec \theta \left( \sec \theta+2 \right)-1\left( \sec \theta+2 \right)=0\] \[\left( \sec \theta+2 \right)\left( 2\sec \theta-1 \right)=0\] \[Either~\sec \theta+2=0,\sec \theta=-2,\cos \theta=-\frac{ 1 }{ 2 }=-\cos \frac{ \pi }{ 3 }=\cos \left( \pi \pm \frac{ \pi }{ 3 } \right)\] \[\theta=\pi \pm \frac{ \pi }{ 3 }\] \[or~2\sec \theta-1=0.\sec \theta=\frac{ 1 }{ 2 },\cos \theta=2\] rejected as\[\left| \cos \theta \right|\le 1\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question