Maths Trig Question

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

    Thanks in advance
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    (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
    Harmonic Form?
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    (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
    Start by changing 4sin(theta)cos(theta) using an identity.

    Then consider an Rcos/Rsin identity.
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    (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?
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    (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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