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nata0808 [166]
3 years ago
11

Solve for 2cos^2theta + 13costheta = -6. theta is >= 0 and < 360 degrees

Mathematics
1 answer:
LiRa [457]3 years ago
7 0
\\ 2cos^2 \theta+13cos \theta=-6 \\ 2cos^2 \theta+13cos \theta+6=0 \\ 2cos^2 \theta+12cos \theta+1cos \theta+6=0 \\ 2cos \theta(cos \theta+6)+1(cos \theta+6)=0  \\ (2cos\theta +1)(cos\theta+6)=0 \\

cos\theta+6=0 \\ cos\theta=-6 \\    \\ cos(x)=|a| \leq 1 \\ |-6| \geq 1 \\  \\


2cos \theta+1=0 \\ 2cos \theta=-1 \\ cos \theta=- \frac{1}{2} \\ \theta= \frac{ 2\pi }{3} +2 \pi k~~~k \in Z \\  \\ 0 \leq \theta \ \textless \ 360^{а}  \\ 0 \leq \theta \ \textless  \ 2 \pi  \\  \theta = \frac{2 \pi }{3}  \\ \theta = \frac{2 \pi }{3} ,  \frac{4 \pi }{3}


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\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0

\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0

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\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}

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It should be easy enough to see that

a_{n=2k} = \dfrac{a_0}{(2k)!}

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so that

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\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}

4 0
3 years ago
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