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Mathematics 26 Online
OpenStudy (anonymous):

solve for theta. cos(2θ) +4 = 5cosθ

OpenStudy (anonymous):

27.5

OpenStudy (anonymous):

how did you do that?

OpenStudy (anonymous):

like the steps

OpenStudy (anonymous):

Use a trig identity to express \(\cos2\theta\) in terms of \(\theta\), then solve the resulting quadratic for \(\cos\theta.\)

OpenStudy (anonymous):

hmmmm let me try that out

OpenStudy (anonymous):

still not grasping it

OpenStudy (anonymous):

Look up an identity for \(\cos2\theta\)!

OpenStudy (anonymous):

cos^2θ - sin^2θ

OpenStudy (anonymous):

OK! Now eliminate \(\sin^2θ\) using \(\sin^2θ=1−\cos^2θ\).

OpenStudy (anonymous):

ok....

OpenStudy (anonymous):

Then substitute back in the original expression and you get a quadratic equation in \(\cos\theta\).

OpenStudy (mertsj):

\[\cos 2x+4=5\cos x\] \[2\cos ^2x-1+4=5\cos \] \[2\cos ^2x-5\cos x+3=0\] \[(2\cos x-3)(\cos x-1)=0\] \[\cos x=\frac{3}{2}, discard\] \[\cos x=1\] x=0

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