1. ## trig functions

solve

sqrt3cosxtanx+cosx where 0 less than or equal to x less than 2pi
and

cos2x-3sinx=2 where -pi is less than or equal to x less than or equal to pi

exactly.

what I tried was

sqrt3cosx*sinx/cosx+cosx=0
sqrt3sinx+cosx=0
then I made each of them equal to 0
sinx=
0/sqrt3
and
cosx=0
But that gives me the wrong answer

And for the other one I tried a similar thing but I couldnt figure it out....
thanks/

2. Originally Posted by brentwoodbc
solve

sqrt3cosxtanx+cosx where 0 less than or equal to x less than 2pi
and

cos2x-3sinx=2 where -pi is less than or equal to x less than or equal to pi

exactly.

what I tried was

sqrt3cosx*sinx/cosx+cosx=0
sqrt3sinx+cosx=0
then I made each of them equal to 0
sinx=
0/sqrt3
and
cosx=0
But that gives me the wrong answer

And for the other one I tried a similar thing but I couldnt figure it out....
thanks/
I can't interpret the first one because you have grouped it funny, and it doesn't seem to be an equation at all!

the second one however..........

Which one should we use? well we probably would like everything to be in terms of sin so........

$\displaystyle 1-2sin^2x+3sinx=2$

$\displaystyle 2sin^2x-3sinx+1=0$

you know how to solve quadratics right? (2x-1)(x-1)?......

3. ya thanks that sort of works I get pi/6, 5pi/6 and pi/2 but the answer is pi/6,5pi/6 and pi/2 but all negatives :/

if the restriction is -pi < x < pi with sine I thought it would be +

also the other one was √3cosx*tanx+cosx=0

4. Originally Posted by brentwoodbc
also the other one was √3cosx*tanx+cosx=0
You mean this?
$\displaystyle \sqrt{3}\cos x \tan x + cos x = 0$

If so, then...

\displaystyle \begin{aligned} \sqrt{3}\cos x \tan x + \cos x &= 0 \\ \cos x(\sqrt{3}\tan x + 1) &= 0 \end{aligned}

Set each factor equal to zero:

\displaystyle \begin{aligned} \cos x &= 0 \\ x &= \frac{\pi}{2}\;\;or\;\;\frac{3\pi}{2} \end{aligned}
or
\displaystyle \begin{aligned} \sqrt{3}\tan x + 1 &= 0 \\ \sqrt{3}\tan x &= -1 \\ \tan x &= -\frac{1}{\sqrt{3}}\;\;or\;\;-\frac{\sqrt{3}}{3} \\ x &= \frac{5\pi}{6}\;\;or\;\;\frac{11\pi}{6} \end{aligned}

01

5. thanks that makes sense, and the way you wrote it was right,
but I have a few problems

1 Is why dont you factor out the cos in the tan "sinx/cosx"
2 (more important) the answer given is 5pi/6 and 11pi/6 and if you graph it you only get the 2 answer's given?
3 for the other problem I had do you know why the answers given are negative? I thought the pi/2 was + and the other two were negative?

6. Originally Posted by VonNemo19
I can't interpret the first one because you have grouped it funny, and it doesn't seem to be an equation at all!

the second one however..........

Which one should we use? well we probably would like everything to be in terms of sin so........

$\displaystyle 1-2sin^2x+3sinx=2$

$\displaystyle 2sin^2x-3sinx+1=0$

you know how to solve quadratics right? (2x-1)(x-1)?......
how did

1-2sin^2x+3sinx=2
become a + ?

7. Originally Posted by brentwoodbc
1 Is why dont you factor out the cos in the tan "sinx/cosx"
I don't understand what you're asking.
2 (more important) the answer given is 5pi/6 and 11pi/6 and if you graph it you only get the 2 answer's given?
Because I forgot to mention that $\displaystyle x = \frac{\pi}{2}\;\;or\;\;\frac{3\pi}{2}$ are extraneous solutions -- if you plug either into the original equation tan x is undefined.

01

8. you jerk

seriously thanks.