## anonymous 4 years ago solve 2secx+tanx=3 on the interval xE[0,2pi]

1. anonymous

this is a pain.

2. anonymous

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

3. anonymous

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

4. anonymous

no that is a mistake

5. anonymous

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

6. anonymous

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

7. anonymous

let me try something else

8. anonymous

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

Find more explanations on OpenStudy