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