Question

solve [0,2pi)
square 12 tanx-2=0

Answer 1

can you wirte it out so that I can understand it, you wrote it kind of abstrusely.

Answer 2

\[\sqrt{12}\tan x - 2 =0\]

Answer 3

so \[\sqrt{12tanx}-2=0\]add two to both sides\[\sqrt{12tanx}=2\]take out 2 from the square root.\[\sqrt{12tanx}=2--->\sqrt{4} \times \sqrt{3tanx}=2--->2\sqrt{3tanx}=2\]divide both sides by 2\[\sqrt{3tanx}=1\]square both sides\[(\sqrt{3tanx})^2=1^2⇒3tanx=1\]divide both sides by 3.\[tanx=1/3\]take inverse tangent of 0.33

Answer 4

I don't have a calculator, sorry.

Answer 5

@shorouq91, is this good?

Answer 6

i'm working out

Answer 7

why\[\sqrt{\tan }\]

Answer 8

Oh, my bad!

Answer 9

no it is\[tanx= \frac{1}{2\sqrt{3}}\]so far so good?

Answer 10

what step i have to used

Answer 11

do you have to show you work?

Answer 12

yes

Answer 13

just give me th \[\sqrt{12}\]

Answer 14

\[\sqrt{12}tanx=2\]simplify the sqrt 12,\[\sqrt{12}tanx=3 ⇒ \sqrt{4} \times \sqrt{3} tanx=2⇒ 2\sqrt{3}tanx=2\]divide both sides by 2, \[\sqrt{3}tanx=1\]

Answer 15

\[\sqrt{3} tanx=1\]divide both sides by sqrt 3

Answer 16

\[tanx=\frac{1}{\sqrt{3}}\]

Answer 17

find what \[\frac{1}{\sqrt{3}}=?\]and take inverse tan of that.

Answer 18

cos/sin
right

Answer 19

The other wAY