Trigi

Solve the following equation for $\theta$, in the interval $0<\theta<360^0$:

$(Sin \theta - 1)(5cos \theta + 3) = 0$

Why can I not do them individually? So...

$Sin \theta - 1 = 0$ $5cos \theta + 3 = 0$

$Sin \theta = 1$ $cos \theta = -\frac {3}{5}$

then proceed from there?

correct

My answers don't correspond with the mark scheme

then you must be doing something else wrong

Anyone?

Anyone?

Sorry. Another question, same range.

$tan \theta = tan \theta (2+3sin \theta)$

$\frac{tan \theta}{tan \theta} = 2 + 3sin \theta$

$Sin \theta = -\frac {1}{3}$

With this I got 2 of the solutions. However, apparently $180^o$ & $360^o$ are solutions to?

This guy is some level of flop.

I know, but all the embarrassing questions will pay off when I get an A.

It's not embarrassing in the least nor are you a flop in anyway.

Cheers!

This will probably be my last question on trig:

$2tan^2x - tan x = 6$
$tan x (2tan x - 1) = 6$
$tan x = 6$ or $2tan x - 1 = 6$

Why would this approach be wrong?

Cheers!

This will probably be my last question on trig:

$2tan^2x - tan x = 6$
$tan x (2tan x - 1) = 6$
$tan x = 6$ or $2tan x - 1 = 6$

Why would this approach be wrong?

You can only factorise stuff and equal it to 0 if it's an equation of them form =0.

E.g, if you had x^2 - 2x = -1

Would you collect that as a quadratic and aolve or would you do x(x-2)= -1 so x = -1 and x -2 = -1.

Does that sound reasonable? What are the correct answers to that quadratic?

Oh, yes. I think I understand it now. I was looking at these trig equations all wrong. I got -116.6, -56.3, 63.4, 123.7 all to 1 dp which corresponds with the answers.

Thanks again!