Trigonometric Equation help

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jessilovin

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Jul 17, 2008
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Find the solutions of the equation that are in the interval [0,2?). Please show steps (especially rearranging the equation to solve it. Thanks for helping me. I know the solutions are 11?/6, ?/2 but I just can't
figure out how to isolate to get those solutions. Thanks again.

1) 1-sint=(?3)cost
 
The easiest way would probably be to plot y = 1 - sin x and y = ?3cos x, then you could just take the solutions off the graph.

Algebraically, we re-arrange the equation to 1 = ?3cos t + sin t

We then assume that ?3cos t + sin t is identically equal to Rcos ( t - a).

Hence, ?3cos t + sin t = Rcos t cos a + Rsin t sin a

Assuming, t = 0 : ?3 = R cos a
Assuming, t = pi/2 : 1 = Rsin a

Hence, solving simultaneously,

(Rsin a) / (Rcos a) = 1/?3 or tan a = 1/?3, a = pi/6

Also, R^2 cos^ a = 3 and R^2 sin^2 a = 1

Adding them togethor we get, R^2 cos^2 a + R^2 sin^2 a = 3 + 1
R^2 (Cos^ a + sin^ a) = 4

as cos^ a + sin^ a = 1,

R^2 = 4, R = 2

Hence, 1 = ?3 cos t + sin t = 2cos (t - pi/6).

And so t - pi/6 = pi/3 or 2pi - pi/3

t = pi/2 or 11pi/6

You could acheive the same result if you assumed that sin t + ?3 cos t = Rsin (t +a ).

Hope this helped!
 
I would square both sides of the equation and then, substitute 1-(sin t)^2 in place of (cos t)^2.
You can then solve the resulting quadratic equation. I went as far as getting sin t = -1/2 or sin t = 1.
When you determine the values of t be sure to check your result(s) back in the original equation. This is necessary because you may have introduced extraneous roots in the squaring process.
 
Hello, jessilovin!

Here's a clever (but obscure) solution . . .

\(\displaystyle \text{Find the solutions of the equation in the interval }[0,2\pi)\)

. . \(\displaystyle 1)\;\;1-\sin t\;=\;\sqrt{3}\cos t\)
Click to expand...

\(\displaystyle \text{We have: }\;\sin t + \sqrt{3}\cos t \;=\;1\)

\(\displaystyle \text{Divide by 2: }\;\frac{1}{2}\sin t + \frac{\sqrt{3}}{2}\cos t \;=\;\frac{1}{2}\)

\(\displaystyle \text{We have: }\;\cos\frac{\pi}{3}\sin t + \sin\frac{\pi}{3}\cos t \;=\;\frac{1}{2}\)

. . \(\displaystyle \text{which equals: }\;\sin\left(t + \frac{\pi}{3}\right) \;=\;\frac{1}{2}\)


\(\displaystyle \text{Then: }\;t + \frac{\pi}{3}\;=\;\left\{\frac{\pi}{6},\:\frac{5\pi}{6}\right\} \quad\Rightarrow\quad t \;=\;\left\{-\frac{\pi}{6},\:\frac{\pi}{2}\right\}\)

. . \(\displaystyle \text{Therefore: }\;t \;=\;\left\{\frac{11\pi}{6},\:\frac{\pi}{2}\right\}\)

 
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