How do you solve: #csc^4x-4csc^2x=0#?

2 Answers
Sean
Apr 28, 2018

Within one period of #0 < x < 2pi#:

#x=pi/6, (5pi)/6, (7pi)/6, (11pi)/6#

Explanation:

.

#csc^4x-4csc^2x=0#

#csc^2x(csc^2x-4)=0#

#csc^2x=0, :. cscx=0, :. 1/sinx=0, :. sinx=oo# Invalid

#csc^2x-4=0, :. csc^2x=4, :. cscx=+-2#

#cscx=2, :. 1/sinx=2, :. sinx=1/2, :. x=pi/6, (5pi)/6#

#cscx=-2, :. 1/sinx=-2, :. sinx=-1/2, :. x=(7pi)/6, (11pi)/6#

Douglas K.
Apr 28, 2018

Given: #csc^4(x)-4csc^2(x)=0#

Factor:

#csc^2(x)(csc(x)-2)(csc(x)+2)=0#

Set each factor equal to 0:

#csc^2(x) = 0#, csc(x) - 2 = 0, csc(x)+2 = 0#

We shall discard the first root because it is outside of the range of the cosecant function, #csc(x) <=-1# and #csc(x) >= 1#; this leaves two factors:

#csc(x) - 2 = 0 and csc(x)+2 = 0#

#csc(x) = 2# and csc(x) = -2#

The cosecant is the reciprocal of the sine:

#sin(x) = 1/2# and #sin(x) = -1/2#

#x = sin^-1(1/2)# and #x = sin^-1(-1/2)#

The primary values are:

#x = pi/6# and #x = -pi/6#

The above primary values repeat at integer multiples of #pi#:

#x = pi/6+ npi and x = -pi/6+ npi, n in ZZ#

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