Question

If `8sinx+3cosx=5` then what is `cot x` ?

Answer 1

`cot x =3/2+-5/4sqrt(3)`

Explanation:

Given:

`8sinx+3cosx=5`

Subtract `3cosx` from both sides to get:

`8sinx=5-3cosx`

Square both sides to get:

`64sin^2x=25-30cosx+9cos^2x`

Note that squaring will not introduce spurious solutions in this case since `sin(-x) = -sin(x)` but `cos(-x) = cos(x)`. So both signs are possible.

Use `cos^2x+sin^2x=1` to reexpress the left hand side and get:

`64-64cos^2x = 25-30cosx+9cos^2x`

Add `64cos^2x-64` to both sides to get:

`0 = 73cos^2x-30cosx-39`

Use the quadratic formula to get:

`cos x = (30+-sqrt((-30)^2-4(73)(-39)))/(2*73)`

`color(white)(cos x) = (30+-sqrt(900+11388))/146`

`color(white)(cos x) = (30+-sqrt(12288))/146`

`color(white)(cos x) = 15/73+-32/73sqrt(3)`

Then from the original equation:

`sin x = (5-3cos x)/8`

So if `cos x = 15/73+32/73sqrt(3)` then:

`sin x = (5-3(15/73+32/73sqrt(3)))/8`

`color(white)(sin x) = 40/73-12/73sqrt(3)`

resulting in:

`cot x = cos x / sin x`

`color(white)(cot x) = (15/73+32/73sqrt(3))/(40/73-12/73sqrt(3))`

`color(white)(cot x) = (15+32sqrt(3))/(40-12sqrt(3))`

`color(white)(cot x) = ((15+32sqrt(3))(10+3sqrt(3)))/(4(10-3sqrt(3))(10+3sqrt(3)))`

`color(white)(cot x) = (150+45sqrt(3)+320sqrt(3)+288)/(4(100-27))`

`color(white)(cot x) = (438+365sqrt(3))/292`

`color(white)(cot x) = 3/2+5/4sqrt(3)`

Similarly, if `cos x = 15/73-32/73sqrt(3)` then:

`cot x = 3/2-5/4sqrt(3)`

Answer 2

`cotx=1/4(6+-5sqrt3), or, 3/2+-5/4sqrt3.`

Explanation:

Given that, `8sinx+3cosx=5......(ast).`

Dividing by `sinx!=0," we get, "8+3cotx=5cscx," &, squaring,"`

`64+48cotx+9cot^2x=25csc^2x=25(1+cot^2x)`

`rArr 16cot^2x-48cotx=39`

Completing square, `16cot^2x-48cotx+36=39+36=75`

`:. (4cotx-6)^2=75 rArr 4cotx-6=+-5sqrt3`

`"Therefore, "cotx=1/4(6+-5sqrt3), or, 3/2+-5/4sqrt3.`

The above soln. was derived on the assumption that, `sinxne0.`

If, `sinx=0,` then, `cosx=+-1,` and sub.ing these in `(ast),`

we get, `+-3=5`, which is impossible. Hence, `sinxne0` holds good.

Enjoy Maths.!