(i) Multiply both sides of
[tex]10x^2 - 19x - 32 = \dfrac3x[/tex]
by x and rearrange terms to get
[tex]10x^3 - 19x^2 - 32 x = 3 \\\\ 10x^3-19x^2-32x-3 = 0[/tex]
The left side factorizes to
[tex](x-3)(x+1)(10x+1) = 0[/tex]
which means x = 3, x = -1, or x = -1/10. We only care about 0 ≤ x ≤ 6, so we take x = 3.
(ii) In the equation
[tex]10\tan\left(\dfrac{2x}3\right) - 19 - 32\cot\left(\dfrac{2x}3\right) = 3\cot^2\left(\dfrac{2x}3\right)[/tex]
notice that multiplying both sides by [tex]\tan\left(\frac{2x}3\right)[/tex] gives the same equation as in (i), but with x swapped out for [tex]\tan\left(\frac{2x}3\right)[/tex] :
[tex]10\tan^2\left(\dfrac{2x}3\right) - 19\tan\left(\dfrac{2x}3\right)- 32 = 3\cot\left(\dfrac{2x}3\right)[/tex]
Then it follows that
[tex]\left(\tan\left(\dfrac{2x}3\right)-3\right) \left(\tan\left(\dfrac{2x}3\right) + 1\right) \left(10\tan\left(\dfrac{2x}3\right)+1\right) = 0 \\\\ \tan\left(\dfrac{2x}3\right)-3 = 0 \text{ or } \tan\left(\dfrac{2x}3\right) + 1 = 0 \text{ or }10\tan\left(\dfrac{2x}3\right) + 1 = 0[/tex]
Solve each case individually.
• Case 1:
[tex]\tan\left(\dfrac{2x}3\right) - 3 = 0 \\\\ \tan\left(\dfrac{2x}3\right) = 3 \\\\ \dfrac{2x}3 = \tan^{-1}(3) + n\pi \\\\ x = \dfrac32\tan^{-1}(3) + \dfrac{3n\pi}2[/tex]
(where n is an integer)
• Case 2:
[tex]\tan\left(\dfrac{2x}3\right)+1 = 0 \\\\ \tan\left(\dfrac{2x}3\right) = -1 \\\\ \dfrac{2x}3 = \tan^{-1}(-1) + n\pi \\\\ \dfrac{2x}3 = -\dfrac\pi4 + n\pi \\\\ x = -\dfrac{3\pi}8 + \dfrac{3n\pi}2[/tex]
• Case 3:
[tex]10\tan\left(\dfrac{2x}3\right)+1 = 0 \\\\ \tan\left(\dfrac{2x}3\right) = -\dfrac1{10} \\\\ \dfrac{2x}3 = \tan^{-1}\left(-\dfrac1{10}\right) + n\pi \\\\ x = -\dfrac32\tan^{-1}\left(\dfrac1{10}\right) + \dfrac{3n\pi}2[/tex]
From here, it's a matter of determining for which n we have 0 ≤ x ≤ 6.
• Case 1: this happens for n = 0, giving x = 3/2 arctan(3).
• Case 2: this happens for n = 1, giving x = 9π/8.
• Case 3: this happens for n = 0, giving x = 3π/2 - 3/2 arctan(1/10).
