First slide

Trigonometric equations

Question

A solution $(x, y) of x^{2} + 2 x \sin x y + 1 = 0 is$

Moderate

A

(1, 0)
B

$(1, 7 π / 2)$
C

$(- 1, - 7 π / 2)$
D

(– 1, 0)

Solution

When $x = 1, 2 + 2 \sin y = 0 \Rightarrow \sin y = - 1$ and $\frac{7 π}{2} = \sin (3 π + \frac{π}{2}) = - 1$
So $(1, 7 π / 2)$ is a solution.
For $x = - 1, \sin y = 1$ which is not satisfied by $y = 7 π / 2 o r 0 .$

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