How do you solve the trigonometric equation \[5\sin x+3\cos x=5\]?

Hint: We are given the equation \[5\sin x+3\cos x=5\]. We have to find the solution of it, we learn what are solution then we will simplify equation to solve, we will use relation between the different ratios we will use the \[{{\sin }^{2}}x+w{{s}^{2}}x=1\]. We will use that in which quadrant which ratio is positive or negative, we also need the knowledge that the sin and cos are periodic functions.

Complete step-by-step solution:
Now we have \[5\sin x+3\cos x=5\]\[\] we have to find its solution using the value of n which will satisfy our equation. As we know that sin and cos are periodic functions then they will have infinitely many solutions. So, we will find the general solution of this equation.
Now, we have
\[{{\sin }^{2}}x+co{{s}^{2}}x=1\]
We subtract $5\,\sin x$ both sides, we get
$3\cos x=5-5\sin x$
Now we will square the sides
So, ${{\left( 3\cos x \right)}^{2}}={{\left( 5-5\sin x \right)}^{2}}$
We get,
\[9{{\cos }^{2}}x=25-50\sin x+25{{\sin }^{2}}x\]
As we know \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\]
So, we get
Using this above we get
$9(1-{{\sin }^{2}}x)= 25-50\sin x+25{{\sin }^{2}}x$
Opening brackets and simplifying by taking all terms to be left, we get
$9-9{{\sin }^{2}}x-25+50\sin x - 25{{\sin }^{2}}x=0$
Simplifying further, we get
$34{{\sin }^{2}}x-50\sin x+16=0$
Now we can see that $34,50$and $16$ has $2$as common as we divide both sides by $2$. So we get
$17{{\sin }^{2}}x-25\sin x+8=0$
Now we will see ${{\sin }^{2}}x$ and change the equation as
$17{{\sin }^{2}}x-25\sin x+8=0$
Now we will use middle term split to factor it, we have
$a=17,\,b=-25\,,\,c=8$
As $17\times 8=a\times c\,\,\,\text{and}\,\,17+8=25$
So,
$\begin{align}
  & 17{{y}^{2}}-25y+8= \\
 & 17{{y}^{2}}-(17+8)y+8=0 \\
& 17{{y}^{2}}-17y-8y+8=0 \\
\end{align}$
So, we get $1$
$=(y-1)(17y-8)=0$
Replace y=sinx, we get
$(\sin x-1)(17 \sin x-8)=0$
So, either $\sin x=1$ on $\sin x=\dfrac{8}{17}$
As we know that $\sin \dfrac{\pi }{2}=1$
So $\sin x=1$
Means $x=\dfrac{\pi }{2}+2x\pi $ x is integer
$2x\pi $ is added because sin is periodic so solution repeat offers $2\pi $
And is $\sin x=\dfrac{8}{17}$
As sin is periodic
So, solution is
$x={{\sin }^{-1}}\left( \dfrac{8}{10} \right)+2\pi x$, x is integer

Note: Remember if we do not give the domain of the function so we will always give a general solution, that stays true for all domains. If we have $\sin x=1$ and write $x=\dfrac{\pi }{2}$ then it will not be fully correct. So, we need to be clear when we need to write the general solution.