Question

picture below again.

Answer 1

Use The same Steps that Sam Used ,Take 3 as a common factor then use trig. identities

Answer 2

okok

Answer 3

this looks very similar to your previous question

remember
\[ \sin^2(\theta) + \cos^2(\theta) = 1\]
this means
\[ \sin^2(\theta) = 1 - \cos^2(\theta) \\ \text{ and } \\\cos^2(\theta) = 1 - \sin^2(\theta)\]

Answer 4

i'm just super clueless when it comes to math. i took it in 11th grade but i never saw any of this

Answer 5

Each time you do a problem, make a note of what did not make sense.
then go the Khan's site, and see if you can find a video that explains it.
Here is "basic trig"
http://www.khanacademy.org/math/trigonometry/basic-trigonometry

but there is more on algebra, square roots, and on and on....

Answer 6

ok thank you very much

Answer 7

in the meantime, can you do this problem ?

Answer 8

the first step is algebra: factor out 3 from (3 - 3cos^2 x)

you "undo" the distributive property
http://www.khanacademy.org/math/arithmetic/multiplication-division/ditributive_property/v/the-distributive-property

you get \(3(1-\cos^2(x)) \)

now use trig to replace the 1-cos^2

Answer 9

next you simplify the square root
\[ \sqrt{3 \cdot \sin^2(x) }\]
see http://www.khanacademy.org/math/algebra/exponent-equations/exponent-properties-algebra/v/simplifying-square-roots

Answer 10

sorry was afk. lemme read this

Answer 11

simplifying the square root is gonna be \[\sqrt{3}\sqrt{\sin^2(x)}\] right?

Answer 12

yes, but you can simplify further
\[ \sqrt{ \sin^2(x)} = \sin(x) \]

Answer 13

you have, so far,
\[ \cos(x) - \sqrt{3} \sin(x) = 0 \]

Answer 14

and that and the other trig function simplify into the 3rd trig function that is missing, right? if you know what i'm trying to say

Answer 15

so cos/sin=tan\[\sqrt{x}\]

Answer 16

?

Answer 17

The definition of tan is sin/cos

if you start with
\[ \cos(x) - \sqrt{3} \sin(x) = 0 \]
and you divide both sides of the equation (all terms) by cos(x) you get
\[ \frac{\cos(x)}{\cos(x)} - \sqrt{3} \frac{\sin(x)}{\cos(x)} = \frac{0}{\cos(x) } \]

now simplify.

Answer 18

what is cos x/ cos x ? anything divided by itself is 1

what is sin x / cos x ?

what is 0/cos(x) 0 divided by anything (except 0) is still 0

Answer 19

so 1-\[\sqrt{3}\tan\] = 0

Answer 20

yes
\[ 1 - \sqrt{3} \tan(x) = 0 \]
we want to "solve" for tan(x)
one way: add \( \sqrt{3} \tan(x) \) to both sides:
\[ 1 - \sqrt{3} \tan(x) + \sqrt{3} \tan(x) = 0 + \sqrt{3} \tan(x)\]

the stuff on the left side simplifies to just 1 (anything plus its negative is zero)
the

Answer 21

you get
\[ 1 = \sqrt{3} \tan(x) \]
divide both sides by \( \sqrt{3} \)
what do you get ?

Answer 22

You should know the sin, cos, and tan of certain special angles
see
http://www.regentsprep.org/Regents/math/algtrig/ATT1/trigANGLES.htm

Answer 23

1/root 3 is what you get, then tan x

Answer 24

so that's the answer?

Answer 25

next you need to know you can add 360º to an angle and you are back where you started (you went all the way around the circle.)
people show this by adding 360*n (n is assumed to be an integer)

you also need to know that the tangent can have 2 different angles that give the same number.

in this case quadrant I and quadrant III both give the same number.

Answer 26

if you have a calculator, you can check the different choices

Answer 27

** 1/root 3 is what you get, then tan x

yes, that is good (so far)

\[ \tan(x) = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \]
people don't like square roots in the denominator, so they often multiply top and bottom by sqrt to get the 2nd form.

now use a table to find what angle that is (it's special, and you are expected to memorize this one)

or use a calculator to find the inverse tangent.