another Q: root(3) sec @ = 2; i got to root(3) sec-2 = 0

[hotlajake]

ok i have this

root(3) sec @ =2

i got to

root (3) sec -2 = 0

what do i do now?

 

[Deleted member 4993]

Re: another question. trig

ok i have this

root(3) sec @ =2

$\displaystyle \sqrt{3} \cdot sec\theta \, = \, 2$

$\displaystyle \sqrt{3} \cdot \frac{1}{cos\theta} \, = \, 2$

$\displaystyle cos\theta \, = \, \frac{\sqrt{3}}{ 2}$

Now go to your favorite picture of unit circle and find out for which angles cos satisfies the given value.

 

[soroban]

Hello, hotlajake!

$\displaystyle \sqrt{3}\sec\theta \:=\:2$

i got to: .$\displaystyle \sqrt{3}\sec\theta -2 \:= \:0$ . . . . why?

Think about what you're doing . . .

You are trying to solve for $\displaystyle \theta$, right?
. . And that means: get $\displaystyle \theta$ alone . . . naked!

\(\displaystyle \text{So we get rid of the }\sqrt{3}\!:\;\;\sec\theta \:=\:\frac{2}{\sqrt3}} \quad\Rightarrow\quad \cos\theta \:=\:\frac{\sqrt{3}}{2}\)

\(\displaystyle \text{Then we get rid of the }\cos\!:\;\;\;\theta \;=\;\cos^{-1}\!\!\left(\frac{\sqrt{3}}{2}\right)\quad\hdots\quad There!\)

$\displaystyle \text{And we're expected to know what angles have a cosine of }\frac{\sqrt{3}}{2}$