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# Maths Trig Question

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

2. (Original post by Fatts13)
Find the solutions of:
2cos2(theta) +4sin(theta)cos(theta) = root 2
in the range 0<=theta<= 2pi

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

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

Then consider an Rcos/Rsin identity.
4. (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?
5. (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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