How do you solve sin x(cosx -1)= 0 over the interval 0 to 2pi?

1 Answer
turksvids
Jan 9, 2018

The solutions to the equation are #x=0#, #x=pi#, and #x=2pi#.

Explanation:

First if #a*b=0# then either #a=0# or #b=0#, so given

#sin(x)(cos(x)-1)=0#

we know that either:

#sin(x)=0# or #cos(x)-1=0#.

For the first equation: #sin(x)=0# happens at #x=0#, #x=pi#, and #x=2pi#. (The Unit Circle is extremely useful for remembering sines and cosines of quadrantal angles.)

For the second equation: #cos(x) -1 =0# gives us #cos(x) = 1#.
The only solutions to this equation on the interval are #x=0# and #x=2pi#, also from the Unit Circle.

So overall the solutions to the equation are #x=0#, #x=pi#, and #x=2pi#.

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