Need help on a trig problem

[Faith54]

Now, I understand problems such as sin2theta=1 up until the point where I have to find all the solutions. For example, sin2theta=90, theta=45. I get that. But where do I go from there? If someone could explain in both degrees and radians (not using a calculator) that would be GREATLY appreciated. Thank you.

 

[Gene]

Sin(2theta) = 90 is wrong. 2theta = 90 is possible and theta does = 45 but sin(anything) is between -1 and 1
The 2 is the number of cycles in 360° or 2pi radians.

 

[soroban]

Hello, Faith54!

Now, I understand problems such as: $\displaystyle \sin(2\theta)\,=\,1$
up until the point where I have to find all the solutions.
For example: $\displaystyle 2\theta\,=\,90^o\;\;\Rightarrow\;\;\theta\,=\,45^o$
I get that.. But where do I go from there?

The problem is that inverse trig functions have multiple answers.

For example: if $\displaystyle \sin\theta\,=\,\frac{1}{2}$, we might conclude: $\displaystyle \theta\,=\,30^o$ . . . and stop.

But there many other angles which satisfy the equation:
. . $\displaystyle 150^o,\,390^o,\,510^o,\,\cdots\;\;$ and: -$\displaystyle 210^o,\,$-$\displaystyle 330^o,\,$-$\displaystyle 570^o,\,$-$\displaystyle 690^o,\,\cdots$

So how do we include all of them?

For this example, there are two basic answers:
. . $\displaystyle \theta\,=\,30^o$ in quadrant 1, and $\displaystyle \theta\,=\,150^o$ in quadrant 2.

All the others are simply: $\displaystyle \;\text{(the basic angle)}\,\pm\,\text{(a multiple of 360}^o)$

The solutions can be written: $\displaystyle \;\theta\:=\:\begin{Bmatrix}30^o\,+\,n\cdot360^o \\ 150^o\,+\,n\cdot360^o\end{Bmatrix}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Back to your problem: $\displaystyle \;\sin(2\theta)\:=\:1$

The basic answer is: $\displaystyle \:2\theta\:=\:90^o$

The general answer is: $\displaystyle \:2\theta\:=\:90^o\,+\,n\cdot360^o\;$ for any integer $\displaystyle n$.

So we have: \(\displaystyle \;2\theta\;=\;\cdots\,-630^o,\.-270^o,\,90^o,\,450^o, \,810^o,\,\cdots\)

Therefore: \(\displaystyle \;\theta\;=\;\cdots\,-315^o,\,-135^o,\,45^o,\,225^o,
\,810^o,\,\cdots\)


If the problem restricts $\displaystyle \theta$ to, say, $\displaystyle 0^o\,\leq\,\theta\,<\,360^o$
. . we just pick the ones in that interval: $\displaystyle \;\theta\,=\,45^o,\,225^o$