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Maths Trig Question

Find the solutions of:
2cos2(theta) +4sin(theta)cos(theta) = root 2
in the range 0<=theta<= 2pi

Thanks in advance :smile:
Reply 1
Avatar for IYGB
IYGB
11
Original post by Fatts13
Find the solutions of:
2cos2(theta) +4sin(theta)cos(theta) = root 2
in the range 0<=theta<= 2pi

Thanks in advance :smile:


Harmonic Form?
Reply 2
Avatar for Notnek
Notnek
21
Original post by Fatts13
Find the solutions of:
2cos2(theta) +4sin(theta)cos(theta) = root 2
in the range 0<=theta<= 2pi

Thanks in advance :smile:

Start by changing 4sin(theta)cos(theta) using an identity.

Then consider an Rcos/Rsin identity.
Reply 3
Avatar for Fatts13
Fatts13
OP
9
Original post by notnek
Start by changing 4sin(theta)cos(theta) using an identity.

Then consider an Rcos/Rsin identity.


Thank You
Going upon what you guys have said...
2cos2(theta) + 4sin(theta)cos(theta)
= 2cos2(theta) + 2sin2(theta)
where A= 2 and B=2
therefore R=root(2^2 + 2^2) = root 8
and alpha = tan^-1 1 = pi/4
therefore root8cos ( theta + pi/4)
root8cos (theta + pi/4) = root 2
cos (theta + pi/4) = root2 / root8
cos (theta + pi/4) = 1/2
shift cos -1 (1/2) = pi/3
therefore the solutions are: pi/12 and 17/12pi
I'm hoping I've done it right?
(edited 7 years ago)
Reply 4
Avatar for Fatts13
Fatts13
OP
9
Original post by IYGB
Harmonic Form?


Thank You Going upon what you guys have said...
2cos2(theta) + 4sin(theta)cos(theta)
= 2cos2(theta) + 2sin2(theta)
where A= 2 and B=2
therefore R=root(2^2 + 2^2) = root 8
and alpha = tan^-1 1 = pi/4
therefore root8cos ( theta + pi/4)
root8cos (theta + pi/4) = root 2
cos (theta + pi/4) = root2 / root8
cos (theta + pi/4) = 1/2
shift cos -1 (1/2) = pi/3
therefore the solutions are: pi/12 and 17/12pi
I'm hoping I've done it right?

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