Question

Trig Equations + General Solutions. (Almost complete, need help with last few lines)

Answer 1

Solve the following equations and give angles between 0 and 360.

\[\sin 2\theta -1 = \cos 2 \theta\]

So far I've completed the following!

\[2 \sin \theta \cos \theta = 2 \cos^2 \theta\]

=\[= \frac{ 2 \sin \theta \cos \theta }{ 2 \cos ^2 \theta } = 1 = \frac{ 2 \sin \theta }{ 2 \cos \theta} = \tan \]

Since tan = 1, tan = 45 degrees,

so the general solution is 180n + 45

However, my book says this is only /half/ the answer. The other solution is 180n + 90, how so?

Thanks in advance!

Answer 2

\(2 \sin \theta \cos \theta = 2 \cos^2 \theta\) can be true for \(cos \theta = 0\) too

Answer 3

In general, try to avoid "cancelling things out".
You lose information when you do that.
Rely more heavily on factoring and other techniques.

\[\large\rm \sin \theta \cos \theta=\cos^2\theta\]Subtraction gives us\[\large\rm 0=\cos^2\theta-\sin \theta \cos \theta\]Pulling a cosine out of each term,\[\large\rm 0=\cos \theta(\cos \theta-\sin \theta)\]

Answer 4

And then apply your Zero-Factor Property,\[\large\rm 0=\cos \theta,\qquad\qquad\qquad 0=(\cos \theta - \sin \theta)\]

Answer 5

@Zepdrix @Irishboy123

Sorry guys! I didn't see your replies earlier.

I see what you mean, thanks you both for your help! c: