Solve $25\cos(\theta - 73.74) = 15$ for $ 0\leq \theta \leq 360$ Solve $25\cos(\theta - 73.74) = 15$ for $0 \leq \theta \leq 360$
There are gaps in my understanding, specifically at the very end of this process. I simplify the above to:
$$ \cos(\theta - 73.74) = \frac{3}{5}$$
$$ \theta = 53.15 $$
If anyone could detail the exact process out, with explanations as to why we do what we do next, I would greatly appreciate it.
 A: You can use the CAST method or just sketch the cosine function to see that there are infinitely many solutions. Then find which ones lie in the interval $0^{\circ}\leq \theta \leq 360^{\circ}.$
We have $$ \cos(\theta - 73.74^{\circ}) = \frac{3}{5}$$
$$ \theta - 73.74^{\circ}= 53.1301^{\circ}+360^{\circ}n $$
$$\theta - 73.74^{\circ}= -53.1301^{\circ}+360^{\circ}k $$
That is
$$ \theta = 126.87^{\circ}+360^{\circ}n $$
$$\theta= 20.61^{\circ}+360^{\circ}k $$
for $n,k\in\mathbb Z$. So choosing $n=0$ for the first and $k=0$ for the second we obtain the two solutions in the required interval to be: $$\theta = 126.87^{\circ} \space \space \text{and}\space \space \theta = 20.61^{\circ}.$$
A: I always draw a graph for these questions. The usual method is to substitute $u = x-73.74$, solve for $u$, and then find the values of $x$ which correspond to the solutions in $u$.
$$$$
A similar method:
First, take a look at the graph, and try to see if you can understand how it addresses your question directly.
I've written on the graph $74^\circ$ instead of $73.74^\circ$ because it would take up too much room on the diagram.
$ \cos(\theta - 73.74) = \frac{3}{5}.$
"Principal value": $\theta_1 - 73.74 = \arccos(\frac{3}{5}) = 53.13...^\circ $
$\implies \theta_1 = 53.13 + 73.74 = 126.87...^\circ,$ and since $0 \leq 126.87...^\circ \leq 360^\circ$, this is the first solution. It is the right red circle on my diagram. Also, from the graph, by symmetry around $73.74^\circ$, the left red circle has value $\theta_2 = 73.74 - (126.87...-73.74) = 20.609...^\circ$ So the solutions are: $\theta_1 = 127^\circ$ and $\theta_2 = 20.6^\circ (3sf)$.

A: The cosine value is positive, the corresponding angle is in the first or the fourth quadrant.
$$0 \le \theta \le 360^\circ$$
$$-73.74^\circ \le \theta-73.74^\circ \le (360-73.74)^\circ$$
Hence we have

*

*$$\theta-73.74^\circ = \pm 53.13^\circ$$
A: [The angle unit is the (decimal) degree]
Write your equation under the form
$$\cos(\theta-73.74)=\cos(53.13)$$
Let us proceed by equivalence.
One must not miss the fact that 2 cosines are equal iff the corresponding angles are equal or opposite (mod 360):
$$\cos(u)=\cos(v) \iff u=v+k 360 \ \text{or} \  u=-v+k 360$$
($k \in \mathbb{Z}$) giving :
$$\theta-73.74=53.13+k 360 \ \text{or} \  \theta-73.74=-53.13+k 360$$
otherwise said:
$$\theta-73.74=53.13+k 360 \ \text{or} \  \theta-73.74=-53.13+k 360$$
$$\theta=126.87+k 360  \ \text{or} \  \theta = 20.61+k 360.$$
As the answers are desired in $[0,360)$, we can drop the $k 360$, giving the
$$\text{Final answer:} \ \theta=126.87  \ \text{or} \  \theta = 20.61.$$
A: A polar plot is perhaps more instructive than a graph.
Draw a circle through origin diameter $5$ cutting x-axis at $r=x=3.,\theta=0. $
Pythagorean triplet sides $(3,4,5)$ are seen in the right triangle when you take a good look.
The circle has equation in polar coordinate form
$$ r= a \cos (\theta- \alpha) ;\; r=3, a=5;\; \alpha= 53.13^{\circ};\;$$
or
$$ @\theta=0, r=3,   @\theta=\alpha, r=5,  @\theta=90^{\circ}, r=4 \quad $$
$$ 5 \cos \alpha= 3,\; 5 \sin \alpha=4,\; 5^2=3^2+4^2. \; $$

