Question

How do you solve this please​

Answer 1

(i) Multiply both sides of

`10x^2 - 19x - 32 = \frac3x`

by x and rearrange terms to get

`10x^3 - 19x^2 - 32 x = 3 \\\\ 10x^3-19x^2-32x-3 = 0`

The left side factorizes to

`(x-3)(x+1)(10x+1) = 0`

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

`10\tan\left(\frac{2x}3\right) - 19 - 32\cot\left(\frac{2x}3\right) = 3\cot^2\left(\frac{2x}3\right)`

notice that multiplying both sides by
`\tan\left(\frac{2x}3\right)` gives the same equation as in (i), but with x swapped out for
`\tan\left(\frac{2x}3\right)` :

`10\tan^2\left(\frac{2x}3\right) - 19\tan\left(\frac{2x}3\right)- 32 = 3\cot\left(\frac{2x}3\right)`

Then it follows that

`\left(\tan\left(\frac{2x}3\right)-3\right) \left(\tan\left(\frac{2x}3\right) + 1\right) \left(10\tan\left(\frac{2x}3\right)+1\right) = 0 \\\\ \tan\left(\frac{2x}3\right)-3 = 0 \text{ or } \tan\left(\frac{2x}3\right) + 1 = 0 \text{ or }10\tan\left(\frac{2x}3\right) + 1 = 0`

Solve each case individually.

• Case 1:

`\tan\left(\frac{2x}3\right) - 3 = 0 \\\\ \tan\left(\frac{2x}3\right) = 3 \\\\ \frac{2x}3 = \tan^(-1)(3) + n\pi \\\\ x = \frac32\tan^(-1)(3) + \frac{3n\pi}2`

(where n is an integer)

• Case 2:

`\tan\left(\frac{2x}3\right)+1 = 0 \\\\ \tan\left(\frac{2x}3\right) = -1 \\\\ \frac{2x}3 = \tan^(-1)(-1) + n\pi \\\\ \frac{2x}3 = -\frac\pi4 + n\pi \\\\ x = -\frac{3\pi}8 + \frac{3n\pi}2`

• Case 3:

`10\tan\left(\frac{2x}3\right)+1 = 0 \\\\ \tan\left(\frac{2x}3\right) = -\frac1{10} \\\\ \frac{2x}3 = \tan^(-1)\left(-\frac1{10}\right) + n\pi \\\\ x = -\frac32\tan^(-1)\left(\frac1{10}\right) + \frac{3n\pi}2`

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).