Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
It is
$3\sin x+4\cos x=2 $
What I did is
$(3\sin x)^2=((2-4\cos x)^2$ to get
$9-9\cos^2x=4-16\cos x+16\cos^2 x$ then I got
$25\cos^2 x-16\cos x-5=0$
However I am having trouble finding the root I tried using the quadratic formula and got x=.866 I am not sure what I do wrong.
There is a fact I could find through my old texts that:
If $a,b$ be real numbers which $a^2+b^2\neq 0$ then $$\exists\alpha\in\mathbb R, \;\; a\sin(x)+b\cos(x)=\sqrt{a^2+b^2}\sin(x+\alpha)$$ I added this, maybe be helpful as a hint.
This works out nicely almost-automatically with groebner basis. Your single equation is equivalent to the two equations
$$s^2+c^2=1$$ $$3s+4c=2$$
The groebner basis in lexicographical order is
$$4c+3s-2 ~~\textrm{and}~~ 25s^2-12s-12$$
So solving the quadratic equation $25s^2-12s-12=0$ gives us the solutions! $$s_{1,2}=\frac{6}{25}\frac{+}{-}\frac{4}{25}\sqrt{21}$$ and $c_{1,2}=(2-3s_{1,2})/4$. Then $x_{1,2}=\arcsin(s_{1,2})=\arccos(c_{1,2})$. However, if you have some software to do groebner basis for you, then it probably already has a solve method to use directly on your first equation.
What you need is:
$$a\sin x+b\cos x=\sqrt{a^2+b^2}\sin(x+\varphi),$$ where the phase shift $\varphi=\arcsin\left({b\over \sqrt{a^2+b^2}}\right)$ when $a>0$.
So, in your situation, you get $$5\sin(x+\varphi)=2\implies x+\varphi=\arcsin(2/5) \implies x=\arcsin(2/5)-\arcsin(4/5)\approx-0.51.$$
I'm not sure if you want a solution or all solutions, but this does the bulk of the work.
I assume that you are just looking for solutions in the interval $[0,2\pi]$. And then what you have done is pretty much correct. So you start with
$$3\sin x+4\cos x=2.$$
And this implies $25\cos^2 x-16\cos x-5=0$.
Let $t = cos(x)$. Then the equation just says $25t^2 - 16t - 5 = 0$. The two solutions are $$\begin{align} t &= \frac{16 \pm\sqrt{16^2 + 500}}{50} \\ &= \frac{16 \pm \sqrt{756}}{50} \\ &= \frac{16 \pm 6\sqrt{21}}{50} \\ &= \frac{8\pm 3\sqrt{21}}{25}\\ &= \begin{cases}0.86990908 \\ -0.229900908\end{cases} \end{align} $$ That is you want to find the solutions to $$ \cos(x) = \begin{cases}0.86990908 \\ -0.229900908\end{cases} $$ By taking $\cos^{-1}$ you can get possible solutions, but when you do that you have to remember that $2\pi$ minus each possible solution is also a solution (to the cosine equation. So you get: $$ x = \begin{cases} 0.51558\quad\text{or}\quad 5.76741 \\ 1.80278\quad\text{or}\quad 4.84952 \end{cases}. $$ In getting your quadratic, you squared both sides, so you have to be careful. We should always go back and check that we got valid solutions. When you do that you realize that only $$ x= 1.80278\quad\text{and}\quad x= 5.76741 $$ are valid solutions.
(All this was of course essentially contained in the comments to the question)
