Question

stuck!

Answer 1

\[\sec^2-2=\tan^2\]

Answer 2

\[\frac{ 1 }{ \cos }-2=\frac{ \sin }{ \cos }\]

Answer 3

\[\frac{ 1 }{ \cos }=\frac{ \sin }{ \cos }+2\]

Answer 4

and now I'm stuck

Answer 5

I think sec^2 becomes 1/cos^2
ditto for tan^2

Answer 6

\[\frac{ 1 }{ \cos }^{2}=\frac{ \sin }{ \cos }^{2}+2\]

Answer 7

are you sure about this question?

I can see
\[ 1 + \tan^2 = \frac{\cos^2}{\cos^2}+\frac{\sin^2}{\cos^2} = \frac{1}{\cos^2}= \sec^2\]

Answer 8

But there is a 2 not a 1 so how would you do that

Answer 9

Here is the original question
Find all solutions in the interval [0, 2π).

sec2x - 2 = tan2x

Answer 10

ooh, you mean sec(2x) not sec^2(x)

Answer 11

no it is sec^2 i copied it wrong.

Answer 12

and tan^2

Answer 13

It looks like no solution
\[ \sec^2(x)-2=\tan^2(x) \]
\[ \sec^2(x)=1+ 1+\tan^2(x) \]
and as shown above 1+tan^2 is sec^2
\[ \sec^2(x)=1+ \sec^2(x) \]
there is no solution.

Answer 14

That is what I was thinking. Thank you!

Answer 15

although we could say
\[ \frac{1}{\cos^2(x) } = \frac{1}{\cos^2(x) }+1 \]
and as long as cos(x) is not zero
\[ 1= 1 + \cos^2(x) \]
\[\cos^2(x)= 0 \]
\[\cos(x)= 0 \]
so we could say as x approaches pi/2 and x-> 3pi/2