Question

solve 2secx+tanx=3 on the interval xE[0,2pi]

Answer 1

this is a pain.

Answer 2

we have
\[\frac{2}{\cos(x)}+\frac{\sin(x)}{\cos(x)}=\frac{2+\sin(x)}{\cos(x)}=3\] so
\[2+\sin(x)=3\cos(x)\]
\[3\cos(x)-\sin(x)=2\]
\

Answer 3

now we have to write
\[3\cos(x)-\sin(x)\] as a single function of sine, we get something like
\[\sqrt{10}\sin(x+\tan^{-1}(3))\]

Answer 4

no that is a mistake

Answer 5

\[-\sqrt{10}\sin(x-\tan^{-1}(3))=2\] i think it is right now

Answer 6

i wonder if there is an easier way, but i don't see it. you can divide by
\[-\sqrt{10}\]and take the inverse sine of the result

Answer 7

let me try something else

Answer 8

no just going around in circles. the above will do it. do you have another method you are supposed to use?