Tìm a[2;3] để a0cos(x+a2)dx=sina
Ta có:  a0cos(x+a2)dx=sinasin(x+a2)|a0=sina
sin(a+a2)sin2a=sina2cos(a2+a2)sina2cosa2
[cos(a2+a2)cosa2]sina2=0
\Leftrightarrow       \left[ \begin{array}{l} \cos (\frac{a}{2} + {a^2}) = \cos \frac{a}{2}\\ \sin \frac{a}{2} = 0 \end{array} \right.          \Leftrightarrow      \left[ \begin{array}{l} \frac{a}{2} + {a^2} = \frac{a}{2} + 2k\pi \\ \frac{a}{2} + {a^2} =  - \frac{a}{2} + 2k\pi    (k \in Z)\\ \frac{a}{2} = k\pi    (loại    vì    a \in  [2;3] \end{array} \right.
\Leftrightarrow    \left[ \begin{array}{l} {a^2} = k\pi \\ {a^2} + a - 2k\pi  = 0 \end{array} \right.   (k \in  Z)   \Leftrightarrow    \left[ \begin{array}{l} {a^2} = k\pi \\ a = \frac{{ - 1 \pm \sqrt {1 + 8k\pi } }}{2} \end{array} \right.   (k \in  Z_+)
\Leftrightarrow    \left[ \begin{array}{l} {a^2} = \sqrt {2\pi } \\ a = \frac{{ - 1 + \sqrt {1 + 8\pi } }}{2} \end{array} \right.   ( vì    k \in  Z_+    và   a \in  [2;3]).

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