Target Mathematics

2.50: Cubic factorisation linked to a trigonometric derivative

0606/22/O/N/17 — Question 10 · 11 marks

(i)
Without using a calculator, solve the equation 6c3−7c2+1=06c^{3} - 7c^{2} + 1 = 0.
[5]
(ii)
It is given that y=tan⁡x+6sin⁡xy = \tan x + 6\sin x. Find dydx\dfrac{\mathrm{d}y}{\mathrm{d}x}.
[2]
(iii)
If dydx=7\dfrac{\mathrm{d}y}{\mathrm{d}x} = 7 show that 6cos⁡3x−7cos⁡2x+1=06\cos^{3} x - 7\cos^{2} x + 1 = 0.
[2]
(iv)
Hence solve the equation dydx=7\dfrac{\mathrm{d}y}{\mathrm{d}x} = 7 for 0⩽x⩽π0 \leqslant x \leqslant \pi radians.
[2]