I am having lots of trouble following my notes , the answer I have is :
$$x= – \left(\arctan{\frac{4}{3}}\right) + \left(\frac{\pi}{2}\right) + 2n\pi,n = 0 \pm 1 \pm 2\dots$$
How I have arrived at this point is very vague to me, I realise this is a fairly long question, therefore If writing it out is too long, please can somebody attach a link to a video or website showing how to do it step by step.
To give an idea of what I am already familiar with
I have been using this to try to get to the answer.
However after getting to $$5\sin(x + 53.1)$$
The next line is
$$5\sin(x + 53.1) = \left(\frac{1}{5}\right)$$
then
$$x + 53.1 = \left(\frac{\pi}{2}\right) n\pi$$
then
$$x= – \left(\arctan{\frac{4}{3}}\right) + \left(\frac{\pi}{2}\right) + 2n\pi $$
It is these three lines of work I am unfamiliar with.
As my tutor has not specified what this kind of question is called I find it difficult to search for online.
Any help is greatly appreciated.
Best Answer
you have $$ \sin x \; \; \cdot \frac{3}{5} + \cos x \; \; \cdot \frac{4}{5} = 1. $$ Since $$ \left( \frac{3}{5} \right)^2 + \left( \frac{4}{5} \right)^2 = \frac{25}{25} = 1, $$ we know there is an angle $\theta$ with $$ \cos \theta = \frac{3}{5} , \; \; \sin \theta = \frac{4}{5} .$$ in fact, $$ \tan \theta = (4/5)/ (3/5) = 4/3, $$ and $$ \theta = \arctan \frac{4}{3}. $$ Then your original is $$ \sin x \cos \theta + \cos x \sin \theta = 1, $$ $$ \sin \left( x + \theta \right) = 1. $$
So far, we have $$ \sin \left( x + \arctan \frac{4}{3} \right) = 1. $$ The sine of something can be equal to $1$ when the thing is $\pi/2,$ or $5 \pi/2,$ or $-3\pi/2,$ in general $\frac{\pi}{2} + 2 n \pi.$ Meaning that the collection of values of $x$ becomes $$ x + \arctan \frac{4}{3} = \frac{\pi}{2} + 2 n \pi$$ with integer $n,$ and $$ x = - \arctan \frac{4}{3} + \frac{\pi}{2} + 2 n \pi$$ At this point i would add the fact that $$ \frac{\pi}{2} - \arctan \frac{4}{3} = \arctan \frac{3}{4}, $$ so that we can write $$ \color{blue}{ x = \arctan \frac{3}{4} + 2 n \pi }$$
Show you how to confirm: given $$ \beta = \arctan \frac{3}{4}, $$ let us use the identity $$ 1 + \tan^2 \beta = \operatorname{sec}^2 \beta, $$ giving $$ 1 + \left( \frac{3}{4} \right)^2 = \operatorname{sec}^2 \beta. $$ Then $$ \frac{25}{16} = \operatorname{sec}^2 \beta, $$ $$ \frac{16}{25} = \cos^2 \beta. $$ Since $\beta$ is in the first quadrant, $$ \cos \beta = \frac{4}{5}, $$ therefore $$ \sin \beta = \frac{3}{5}. $$ Then $$ x = \beta + 2 n \pi, $$ $$ \cos x = \frac{4}{5}, $$ $$ \sin x = \frac{3}{5}. $$ $$ 3 \sin x + 4 \cos x = 3 \frac{3}{5} + 4 \frac{4}{5} = \frac{25}{5} = 5 $$