First slide

Trigonometric equations

Question

If $3 \sin x + 4 \cos a x = 7$ has at least one solution then the possible values of a

Difficult

A

$\frac{4 m}{4 n + 1}, m, n \in Z$
B

$(4 n + 1) \frac{π}{2}, n \in Z$
C

$\frac{m}{4 n + 1}, m, n \in Z$
D

$\frac{4 n + 1}{4 m}, m, n \in Z$

Solution

$\begin{array}{l} W e h a v e \sin x = 1, \cos a x = 1 \\ \Rightarrow a x = 2 m π, x = (4 n + 1) \frac{π}{2}, m, n \in Z \end{array}$
$\Rightarrow (4 n + 1) \frac{π}{2} = \frac{2 m π}{a} \Rightarrow a = \frac{4 m}{4 n + 1}, m, n \in z$

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