Question

Determine all of the non-permissible values for x in the following expressions

cot(x)/2sin^2(x)-1

Answer 1

A non-permissable value is any value that will make the expression false. In this case, it would be any value that would make the denominator equal to zero. Can you solve\[2\sin ^{2}(x)-1=0\]

Answer 2

would it be...

\[2\sin^2(x) -1 = 0\]
\[2\sin^2(x)=1\]
\[\sin^2(x)=1/2\]
\[\sin(x)=(1/2)^2\]
\[x=\sin^{-1} ((1/2)^2)\]

Answer 3

You cannot do your fourth step. Here's how I would have simplified it:
\[2\sin ^{2}(x)-1=0\]\[2\sin ^{2}(x)-(\cos ^{2}(x)+\sin ^{2}(x))=0\]\[2\sin ^{2}(x)-\cos ^{2}(x)-\sin ^{2}(x)=0\]\[\sin ^{2}(x)-\cos ^{2}(x)=0\]At this point, you can convert the left side to something simpler. Do you know what that is?

Answer 4

I'm afraid I don't

Answer 5

Alright. Some of your best friends when simplifying trig are the double and half angle formulas (especially in later math classes). In this case, the double angle formula comes in extremely handy. The double angle formula states:
\[\cos(2x)=\cos ^{2}(x)-\sin ^{2}(x)\] Can you see where this can be substituted into the equation?