# Math Help - Need help on these...phew hard one..

1. ## Need help on these...phew hard one..

These I need help on now..they're screwing with my mind!

Solve the equation for solutions in the interval [0, 360). Round to the nearest degree.

sin 2theta =cos theta

sin 2theta= -1/2

sqrt3 sec2theta =2

Solve the equation for x

y=8 cos 3x

Solve the equation for exact solutions.

arcsin(y-pi/6) = pi/3

arctanx = arcsin 7/25

Solve the equation.

arcsin 2x +2 arccos x=pi

2. Originally Posted by Sinsane
These I need help on now..they're screwing with my mind!

Solve the equation for solutions in the interval [0, 360). Round to the nearest degree.

sin 2theta =cos theta

sin 2theta= -1/2
$\sin 2\theta=2\sin\theta\cos\theta$

This answers first two question... Try it..

3. I dont get it...could someone use that too figure it out?

4. NOTE THAT THIS IS A INCOMPLETE SOLUTION AND MAY MISLEAD ANYBODY....SO PLEASE OVERLOOK AT THIS ONE...

Originally Posted by Sinsane
I dont get it...could someone use that too figure it out?
$\sin2\theta=2\sin\theta\cos\theta$

1) $\sin 2\theta=\cos\theta$

$\Rightarrow 2\sin\theta\cos\theta=\cos\theta$

$\Rightarrow 2\sin\theta=1$

$\Rightarrow \sin\theta=\frac1{2}$

$\Rightarrow \theta=30^0$

5. Thanks so much..Now I understand that one. I didnt know that cos and sin could just appear somewhere..haha

6. Originally Posted by great_math
$\sin2\theta=2\sin\theta\cos\theta$

1) $\sin 2\theta=\cos\theta$

$\Rightarrow 2\sin\theta\cos\theta=\cos\theta$

$\Rightarrow 2\sin\theta=1$

$\Rightarrow \sin\theta=\frac1{2}$

$\Rightarrow \theta=30^0$
NO!

Several corrections:

$\Rightarrow 2\sin\theta\cos\theta=\cos\theta$

$\Rightarrow \cos \theta (2\sin \theta - 1) = 0$

Therefore either

$2 \sin \theta - 1 = 0 \Rightarrow \sin \theta = \frac{1}{2}$

OR ${\color{red}\cos \theta = 0}$.

Furthermore, another solution to ${\color{red}\sin \theta = \frac{1}{2}}$ over the interval [0, 360) is ${\color{red}\theta = 150^0}$.

7. Originally Posted by mr fantastic
NO!

Several corrections:

$\Rightarrow 2\sin\theta\cos\theta=\cos\theta$

$\Rightarrow \cos \theta (2\sin \theta - 1) = 0$

Therefore either

$2 \sin \theta - 1 = 0 \Rightarrow \sin \theta = \frac{1}{2}$

OR ${\color{red}\cos \theta = 0}$.

Furthermore, another solution to ${\color{red}\sin \theta = \frac{1}{2}}$ over the interval [0, 360) is ${\color{red}\theta = 150^0}$.
I did not give the complete solution...
I just showed the way..
With my editing i accidentally deleted that proceed now to find other values of $\theta$

I am extremely sorry for this inconvenience...

8. Originally Posted by great_math
I did not give the complete solution...
I just showed the way..
With my editing i accidentally deleted that proceed now to find other values of $\theta$

I am extremely sorry for this inconvenience...
When students are faced with solving equations of the form $x^2 = x$ many have the misconception that you can simply divide by the common factor of x and get x = 1 as the answer. This approach is wrong as it ignores the possibility of x=0.

It's my opinion that from the OP's point of view your reply, unfortunately, reads as a complete solution where this misconception is reinforced. It's unfortunate that your editing also suggested that there was only one solution as this is also a common misconception among novices.

The academic merit of a reply can only be judged on what ends up being posted. If there's a likelihood that the OP will be confused or get the wrong impression from the reply, or if the reply contains errors, then it's the responsibility of all members, not just Moderators, to point this out.

I myself have been on the receiving end of this process and wouldn't want it any other way.