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Mathematics 15 Online
OpenStudy (k8lyn911):

TRIGONOMETRY: Solve the equation for all degree solutions and if 0° ≤ θ < 360°. sqrt(3) tan theta - 1 = 0

OpenStudy (k8lyn911):

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

hartnn (hartnn):

can you isolate tan theta first ? start by adding 1 on both sides :)

zepdrix (zepdrix):

\(\Large\bf \color{#008353}{\text{Welcome to OpenStudy! :)}}\)

OpenStudy (k8lyn911):

Yes, but I can't figure out what to do after that. \[\tan \theta = 1/\sqrt{3}\]

hartnn (hartnn):

there is a standard angle for which tan theta is 1/sqrt 3 do u have unit circle with you ?

OpenStudy (k8lyn911):

I pulled up the unit circle, but I don't see \[1/\sqrt{3}\] anywhere on it.

hartnn (hartnn):

because you only see sin and cos in unit circle... and tan = sin/cos so search for the angles where ratio is 1/sqrt 3

hartnn (hartnn):

y co-ordinate/ x co-ordinate should equal 1/sqrt 3

OpenStudy (k8lyn911):

I already had one pulled up with tan on it. Oh, but I didn't simplify. \[1/\sqrt{3} = \sqrt{3}/3\] right?

hartnn (hartnn):

yes.

hartnn (hartnn):

could u find the angles ? try pi/6 ?

hartnn (hartnn):

sin pi/6 = ...? cos pi/6 =...? tan pi/6 = sinpi/6 / cos pi/6 = ... ?

hartnn (hartnn):

in degrees, pi/6 is 30 degrees

OpenStudy (k8lyn911):

I have the angles: 30, 60, and 240. But my homework is telling me that I'm wrong.

hartnn (hartnn):

how come 60 and 240 ? did you look for 210 ?

OpenStudy (k8lyn911):

Oops, nevermind! Wrong column. I'll fix it.

OpenStudy (k8lyn911):

Okay, I got it this time! Thanks! :)

hartnn (hartnn):

welcome ^_^

hartnn (hartnn):

and again \(\Huge \mathcal{\text{Welcome To OpenStudy}\ddot\smile} \) if u need any help browsing this site, you can ask me :)

OpenStudy (k8lyn911):

Okie dokie. Thanks again! :)

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