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

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)#

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.!