q2) a) write the equation cos2x + 8cosx+9=0 in terms of cosx and show that for cosx it has equal roots

q2b) show that there are no real roots for x.

for q2 i have tried to do it but i get upto the bit 2(cosx+2)(2cosx+2)=0 and i don't know what to do next.

  1. Rewrite the equation with cos x as the variable.
    2 cos^2x -1 + 8 cosx +9 = 0
    2cos^2x +8 cosx +8 = 0
    cos^2x +4x +4 = 0
    (cosx +2)^2 = 0

    Since cosx cannot equal -2 (for real values of x), there are no real solutions. Quadratic equations that can be written as the square of a monomial factor have equal roots.

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  2. actually it should have been
    2cos^2 x - 1 + 8cosx + 9 = 0
    2cos^2 x + 8cosx + 8 = 0
    cos^2 x + 4x + 4 = 0

    so we have shown that for cosx it has equal roots. (the factor cosx+2 appears twice)

    for the second part,
    cosx + 2 = 0
    cosx = -2

    but for all real x, -1 ≤ cosx ≤ 1

    so cosx = -2 is outside that domain, and there are no real roots.

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