A trigonometric equation So I have a solution $x = 5° + n20°$ for a trigonometric equation and I want to find the exact same solutions, but in different form.
Here is what I tried:
$x=−115°+m20°$ 
$x=25°+m40°$ and $x=85°+m40°$  
$x=−45°+m20°$ 
$x=5°+m40°$ and $x=−35°+m40°$ 
Are my solutions correct?
 A: $$x = 5° + n20°\\=5° + (n-6+6)20°\\=
5° -120°+ (n+6)20°\\=
-115° -\underbrace{(n+6)}_{m}20°\\=-115°+m20°$$other choices are not in the form 
.specially $$x=−45°+m20° $$ is wrong because $$x = 5° + n20°\\=x = 5° -40+ m20° \to x = -35° + m20°$$
A: $$x=\frac {\pi}{36}+n\frac {2\pi}{18} $$
to get all solutions , $n $ should take its values in $$\{0,1,2,..17\} $$
A: Technically, $x = 5^\circ + n20^\circ$ is itself an equation.
If we have the condition that $n$ is an integer,
then the equation has a countably infinite number of solutions $(x,n).$
Given the trigonometric equation,
$$ \sin(-5^\circ + 9x) = 0, $$
for example,
we could write the solution set $S$ of this equation as
$$ S = \{ x \mid x = 5^\circ + n20^\circ, n \in \mathbb Z\}. $$
Slightly less formally, 
$$S = \{\ldots,-35^\circ, -15^\circ, 5^\circ, 25^\circ, 45^\circ,\ldots\}.$$
Since $n - 6$ is an integer if and only if $n$ is, 
we can also write the solution set as 
$$ S=\{ x \mid x = 5^\circ + (n-6)20^\circ, n \in \mathbb Z\}. $$
Since $5^\circ + (n-6)20^\circ = -115^\circ + n20^\circ,$
your form $x = -115^\circ + m20^\circ$ (where $m$ is an integer)
appears to be OK.
If $m$ is an integer, however, there is no solution of 
$x = -45^\circ + m20^\circ$ for which $x = 5^\circ.$
In fact, for $m = 2$ we have $x = -5^\circ$ and for
$m = 3$ we have $x = 15^\circ.$ So
$$S \neq \{ x \mid x = -45^\circ + m20^\circ, m \in \mathbb Z\}. $$
If you form one subset of solutions for even values of $n$ and one for odd values of $n$ you get
\begin{align}
S &= \{ x \mid x = 5^\circ + (2k)20^\circ, k \in \mathbb Z\}
\cup \{ x \mid x = 5^\circ + (2k+1)20^\circ, k \in \mathbb Z\} \\
&= \{ x \mid x = 5^\circ + k40^\circ, k \in \mathbb Z\}
\cup \{ x \mid x = 25^\circ + k40^\circ, k \in \mathbb Z\}.
\end{align}
If you replace $k$ with $m+2$ in the first subset and $m$ in the second, you get
\begin{align}
S &= \{ x \mid x = 85^\circ + m40^\circ, m \in \mathbb Z\}
\cup \{ x \mid x = 25^\circ + m40^\circ, m \in \mathbb Z\} \\
&= \{ x \mid x = 25^\circ + m40^\circ \text{ or } 
             x = 85^\circ + m40^\circ, m \in \mathbb Z\}.
\end{align}
That's a lot like your second alternative, but
note that the logical operator to combine the two conditions is "or," not "and." The statement 
"$x = 25^\circ + m40^\circ$ and $x = 85^\circ + m40^\circ$"
has no solutions at all.
The condition $x=5^\circ+m40^\circ$ or $x=−35^\circ+m40^\circ$ 
is incomplete because it is the "even" case of $n$ twice,
and is missing all the "odd" cases; for example, it cannot give
you the solution $x = -15^\circ.$
