SOLUTION: Please help me solve this problem:
Find the solutions on the interval [0,2pi).
{{{sqrt(2)cosxsinx-cosx=0}}}
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-> SOLUTION: Please help me solve this problem:
Find the solutions on the interval [0,2pi).
{{{sqrt(2)cosxsinx-cosx=0}}}
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You can put this solution on YOUR website! factor out the cos(x)
you get cos(x)[sqrt(2)sin(x)-1)]=0
from this it follows that either cos(x)= 0 or [sqrt(2)sin(x)-1)] = 0
if cos(x)=0, then in the given interval, you should get two solutions, i'll leave it to you to find them.
if [sqrt(2)sin(x)-1)] = 0 then x = sinInverse(1/sqrt(2)), and once again you'll get two solutions
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