solving a trig equation

**200001** · February 15th 2010, 10:17 PM

hello

I am thinking this should factorise as the difference of 2 squares or be converted using trig identities

I just keep hitting dead ends though

Any help would be appreciated

**Opalg** · February 16th 2010, 12:12 AM

Originally Posted by 200001

hello

I am thinking this should factorise as the difference of 2 squares or be converted using trig identities

I just keep hitting dead ends though

Any help would be appreciated

Divide through by $\cos^2\phi$ . Then you'll have a quadratic equation for $\tan\phi$ .

**200001** · February 16th 2010, 12:26 AM

Great, that sorts it
how do i decide on future problems what to divide by?

**Soroban** · February 16th 2010, 05:13 AM

Hello, 200001!

This is not a difference of two squares . . .

$3\sin^2\!\theta - \sin\theta\cos\theta - 4\cos^2\!\theta \:=\:0$

Factor: . $(\sin\theta + \cos\theta)(3\sin\theta - 4\cos\theta) \:=\:0$

And we have two equations to solve:

$\sin\theta \;+\; \cos\theta \:=\:0 \quad\Rightarrow\quad \sin\theta \:=\:-\cos\theta \quad\Rightarrow\quad \frac{\sin\theta}{\cos\theta} \:=\:-1 \quad\Rightarrow\quad \tan\theta \:=\:-1$
. . Hence: . $\theta \:=\:\arctan(-1)$

$3\sin\theta \;-\; 4\cos\theta \:=\:0 \quad\Rightarrow\quad 3\sin\theta \:=\:4\cos\theta \quad\Rightarrow\quad \frac{\sin\theta}{\cos\theta} \:=\:\frac{4}{3} \quad\Rightarrow\quad \tan\theta \:=\:\frac{4}{3}$
. . Hence: . $\theta \:=\:\arctan\left(\tfrac{4}{3}\right)$

**Dotdash13** · February 17th 2010, 05:29 PM

Originally Posted by 200001

Great, that sorts it
how do i decide on future problems what to divide by?

With fairly basic trig, you pretty much have less than 10 options total, and you can usually rule several of those out just by looking at what trig functions are being used. for example, in your problem you could have also divided by $sin^2$ , but you would not have really wanted to use something like secant to tangent. You learn mostly with practice.

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