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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. enter image description here

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.

Aztec warrior
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2 Answers2

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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 $$ x = \beta + 2 n \pi, $$ $$ \cos x = \frac{4}{5}, $$ $$ \sin x = \frac{3}{5}. $$ $$ 3 \sin x + 4 \cos x = 3 \frac{3}{5} + 4 \frac{4}{5} = \frac{25}{5} = 5 $$

Will Jagy
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You have $\frac 35\sin x+\frac 45\cos x=1$ which is $\sin x\cos \alpha+\cos x\sin \alpha=1$ where $\alpha$ is the angle in the figure below whose tangent is equal to $\frac43$.

enter image description here

Furthermore you know that $\sin x\cos \alpha+\cos x\sin \alpha=\sin (\alpha+x)$

You have then $\alpha+x=\frac {\pi}{2}+2n\pi$ and $x= -tan^{-1}(4/3) + (π/2) + 2nπ$

Piquito
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