URGENT - Trigonometry - Identities and Proofs

Okay, today, I find myself utterly dumbfounded by these three questions -
Write a proof for -
2/(sqrt(3)cos(x) + sin(x))= sec((pi/6)-x)
Solve the following equation -
2sin(2x) - 2sin(x) + 2(sqrt(3)cos(x)) - sqrt(3) = 0
Find all solutions (exact) to the equation -
sin^2(x)cos^2(x) = (2 - sqrt(2))/16
Any Help at all will most definitely be appreciated!!!

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  1. In proving identities the general rule is to start with the more complicated side and, by mathematical manipulation, show that it is equal to the other side.
    (even though it may not look it, in your problem the rights side (RS) is more complicated, so ....

    RS = 1/cos(π/6-x)
    = 1/cos(30°-x) , it makes no difference if you work in degrees or radians
    = 1/(coss30cosx + sin30sinx)
    = 1/((√3/2)cosx + (1/2)sinx)
    = 1/((√3cosx + sinx)/2)
    = 2/(√3cosx + sinx)
    = LS

    Do either solve or prove these type of equations, you have to be familiar with the basic trig relationships, such as
    cos(A±B) = cosAcosB -/+ sinAsinB, which I used above.
    I also needed to know the trig ratios of the standard 30-60-90 right-angled triangle.

    in the first of your "solve equation" we will need
    sin(2A) = 2sinAcosA

    2sin(2x) - 2sinx + 2√3cosx - √3 = 0
    4sinxcosx - 2sinx +2√3cosx - √3 = 0
    common factors
    2sinx(2cosx - 1) + √3(2cosx - 1) = 0
    (2cosx - 1)(2sinx + √3) = 0
    2cosx-1 = 0 or 2sinx + √3 = 0
    cosx = 1/2 or sinx = -√3/2
    the last of these is not possible since the sine and the cosine fall between -1 and +1

    so cosx = 1/2
    from the 30-60-90 triangle , we know cos 60° = 1/2
    and by the CAST rule, the cosine is positive in I and IV
    so x = 60° or x = 300°
    or
    x = π/3 or x = 5π/3 radians.


    last one:
    sin^2x cos^2x = (2-√2)/16
    take √ of both sides
    sinxcosx = ±√(2-√2)/4
    4sinxcosx = ±√(2-√2
    2sin 2x = ±√(2-√2)
    sin2x = ±√(2-√2)/2

    so 2x is in any of the 4 quadrants
    using my calculator , I evaluated the right side, then by "inverse sin", found
    2x = 22.5° ( ahh, 1/2 of the 45, one of the standard angles)
    so 2x = 22.5 or 157.5 or 202.5 or 337.5
    making
    x = 11.25 , 78.75 , 101.25 or 168.75

    since the period of sin2x , which gave us our answer, is 180/2 or 90°, adding or subtracting multiples of 90 to any of the above answers will produce a new answer.
    e.g. 78.75+90= 168.75° will also be an answer.

    let's test that:
    sin^2 (168.75) + cos^2(168.75)
    = .036611651 by calculator
    RS = (2-√2)/16 = .036611652 , not bad eh?

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    posted by Reiny
  2. In the second equation I rejected
    sinx = -√3/2
    I shouln't have since that value lies between -1 and 1
    from the above we also get
    x = 120° or x =240°
    or
    x = 2π/3 or x = 4π/3 rads

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    posted by Reiny

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