Solve for β to the nearest tenth of a radian for [0,2pi): 2+tanβ = cotβ.
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pythagorean314159
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What are your thoughts? What have you tried? How far have you gotten? Where are you stuck?
For instance, you took the hint provided earlier, then multiplied through to clear the fraction, found that you had a quadratic, and... then what?
Please be complete. Thank you!
What are your thoughts? What have you tried? How far have you gotten? Where are you stuck?
For instance, you took the hint provided earlier, then multiplied through to clear the fraction, found that you had a quadratic, and... then what?
Please be complete. Thank you!
Thanks for the hint, I did get as far as the third line (below) and knew I needed the quadratic formula but then had trouble applying it correctly.
Had help from friend in class, to be honest.. but ended up with these solutions (?)
2+tanβ=cotβ
2+tanβ= 1/tanβ
2tanβ+tan2β=1
tan2β+2tanβ−1=0
tanβ= −2±√22−4.1.(−1)/ 2.1
tanβ=0.4142 & −2.4142
β= tan^−1(0.4142) &tan^-1(−2.4142)
So β rounds to 0.4 and -1.2
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Well, the first thing I noticed was that your work was hard to follow, due to you use of poor notation (for instance, tan2β instead of tan^2(β)) and missing grouping symbols. I believe this was what you did. You started with:
\(\displaystyle 2+tan(\beta)=cot(\beta) \implies 2+tan(\beta)=\dfrac{1}{tan(\beta)}\)
You multiplied through by \(\displaystyle tan(\beta)\) and brought the 1 to the other side, leaving a quadratic:
\(\displaystyle 2tan(\beta)+tan^2(\beta)=1 \implies 2tan(\beta)+tan^2(\beta)-1=0\)
You then solved the quadratic:
\(\displaystyle tan(\beta)=\dfrac{-2 \pm \sqrt{2^2-4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \implies tan(\beta)=\dfrac{-2 \pm \sqrt{8}}{2}\)
Simplifying this further left you with the solutions of \(\displaystyle tan(\beta) \approx 0.4\) and \(\displaystyle tan(\beta) \approx 1.2\)
You seem unsure of your answers. If you're ever unsure of an answer, you can always plug it back in to the original equation to see if it checks. This is also advisable because it allows you weed out extraneous "invalid" solutions. So let's check your answers.
\(\displaystyle tan(\beta)=0.4\) gives:
\(\displaystyle 2+tan(0.4)=cot(0.4)\)
\(\displaystyle 2+0.423=2.365\)
\(\displaystyle tan(\beta)=-1.2\) gives:
\(\displaystyle 2+tan(-1.2)=cot(-1.2)\)
\(\displaystyle 2-2.572=-0.389\)
Hm. Neither of these equations are true. But remember that you rounded heavily, as per the stipulations of the problem. The values are close though, so let's use the exact values and see what happens. The original values that the quadratic equation spat out are:
\(\displaystyle tan(\beta)=\sqrt{2}-1\) and \(\displaystyle tan(\beta)=-\sqrt{2}-1\)
Plugging in \(\displaystyle tan(\beta)=\sqrt{2}-1\) gives:
\(\displaystyle 2+(\sqrt{2}-1)=\dfrac{1}{\sqrt{2}-1}\)
Multiplying by the conjugate and some simplification leaves:
\(\displaystyle 1+\sqrt{2}=\sqrt{2}+1\)
That checks out. Now you try checking your other answer. What do you find? Is it an extraneous solution? If so, why?
\(\displaystyle 2+tan(\beta)=cot(\beta) \implies 2+tan(\beta)=\dfrac{1}{tan(\beta)}\)
You multiplied through by \(\displaystyle tan(\beta)\) and brought the 1 to the other side, leaving a quadratic:
\(\displaystyle 2tan(\beta)+tan^2(\beta)=1 \implies 2tan(\beta)+tan^2(\beta)-1=0\)
You then solved the quadratic:
\(\displaystyle tan(\beta)=\dfrac{-2 \pm \sqrt{2^2-4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \implies tan(\beta)=\dfrac{-2 \pm \sqrt{8}}{2}\)
Simplifying this further left you with the solutions of \(\displaystyle tan(\beta) \approx 0.4\) and \(\displaystyle tan(\beta) \approx 1.2\)
You seem unsure of your answers. If you're ever unsure of an answer, you can always plug it back in to the original equation to see if it checks. This is also advisable because it allows you weed out extraneous "invalid" solutions. So let's check your answers.
\(\displaystyle tan(\beta)=0.4\) gives:
\(\displaystyle 2+tan(0.4)=cot(0.4)\)
\(\displaystyle 2+0.423=2.365\)
\(\displaystyle tan(\beta)=-1.2\) gives:
\(\displaystyle 2+tan(-1.2)=cot(-1.2)\)
\(\displaystyle 2-2.572=-0.389\)
Hm. Neither of these equations are true. But remember that you rounded heavily, as per the stipulations of the problem. The values are close though, so let's use the exact values and see what happens. The original values that the quadratic equation spat out are:
\(\displaystyle tan(\beta)=\sqrt{2}-1\) and \(\displaystyle tan(\beta)=-\sqrt{2}-1\)
Plugging in \(\displaystyle tan(\beta)=\sqrt{2}-1\) gives:
\(\displaystyle 2+(\sqrt{2}-1)=\dfrac{1}{\sqrt{2}-1}\)
Multiplying by the conjugate and some simplification leaves:
\(\displaystyle 1+\sqrt{2}=\sqrt{2}+1\)
That checks out. Now you try checking your other answer. What do you find? Is it an extraneous solution? If so, why?
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