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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.

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    $\color{maroon}{\cos}( x)=.866$ (approximately; which is a bit off)... The variable in your quadratic is $\cos x$, not $x$.2012-12-15
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    You are correct x is closer to .8699092012-12-15
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    But I cannot seem to get the right answer.2012-12-15
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    $x$ has two values $0.86990908339, -0.22990908339$2012-12-15
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    yes because it plus and minus in the quadratic formula.2012-12-15
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    What you get is $$\cos x=\frac{8\pm3\sqrt{21}}{25}$$ So $$x=\arccos\left(\frac{8\pm3\sqrt{21}}{25}\right)$$2012-12-15
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    Alternately, divide through by $5=\sqrt{3^2+4^2}$, and let $\phi$ be the angle whose cosine is $\frac{3}{5}$ and whose sine is $\frac{4}{5}$.2012-12-15
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    I did arccos(.87) and I got .5155 but on the answer key of my book it says the correct answer is 1.80,5.762012-12-15
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    @FernandoMartinez: Did you check whether the answer is correct?2012-12-15
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    Taking the inverse cosine will give you the value of $x$ in the interval $[0,\pi]$ that satisfies the quadratic equation. In this case, you obtain $x\approx .516\,\text{rad}$. But $\cos(x)=\cos(-x)$, so another solution to the quadratic is $x\approx -.516\approx 5.76\,\text{rad}$. Now, you squared both sides of your original equation, so you need to check if these answers satisfy the original equation (before you squared both sides). The former works, the latter doesn't (then $2-4\cos(x)$ is negative). And, don't forget about the other root of the quadratic.2012-12-15
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    Oh I see now the other solution works. thanks.2012-12-15

4 Answers 4