How do you factor this equation in the interval [0,360)? cos^2(x)sin(x) = sin(x)

2 Answers
Sean
Feb 21, 2018

Please see below.

Explanation:

.

#cos^2xsinx=sinx#

#cos^2xsinx-sinx=0#

#sinx(cos^2x-1)=0#

#sinx=0, :. x=0^@, 180^@, 360^@#

#cos^2x-1=0#

#cos^2x=1#

#cosx=+-1#

#cosx=1, :. x=0^@, 360^@#

#cosx=-1, :. x=180^@#

Jim G.
Feb 21, 2018

#x=0^@,180^@#

Explanation:

#"rearrange and equate to zero"#

#rArrcos^2sinx-sinx=0#

#"take out a common factor of "sinx#

#rArrsinx(cos^2x-1)=0#

#"equate each factor to zero and solve for x"#

#sinx=0rArrx=0^@.180^@,cancel(360^@)#

#cos^2x-1=0rArrcos^2x=1rArrcosx=+-1#

#cosx=1rArrx=0^@,cancel(360^@)#

#cosx=-1rArrx=180^@#

#"putting the solutions together gives"#

#x=0^@,180^@x in[0,360)#

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