First slide

Trigonometric equations

Question

For the equation
$1 - 2 x - x^{2} = \tan^{2} (x + y) + \cot^{2} (x + y)$

Moderate

A

Exactly one value of $x$ exists
B

Exactly two values of $x$ exists
C

$y = - 1 + n π + \frac{π}{4}, n \in z$
D

$y = 1 + n π + \frac{π}{4}, n \in z$

Solution

$\begin{array}{l} 1 - 2 x - x^{2} = \tan^{2} (x + y) + \cot^{2} (x + y) \\ \Rightarrow 1 - 2 x - x^{2} - 2 = \tan^{2} (x + y) + \cot^{2} (x + y) - 2 \\ \Rightarrow - {(x + 1)}^{2} = {[\tan (x + y) - \cot (x + y)]}^{2} \\ N o w L . Η . S \leq 0 a n d R . Η . δ \geq 0 \\ \Rightarrow - {(x + 1)}^{2} = {[\tan (x + y) - \cot (x + y)]}^{2} = 0 \\ ⇒ x = - 1 a n d \tan (x + y) = \cot (x + y) \\ ⇒ x = - 1 a n d \tan^{2} (- 1 + y) = 1 \\ \Rightarrow x = - 1 a n d - 1 + y = n π \pm \frac{π}{4} : n \in z \\ \Rightarrow y = 1 + n π \pm \frac{π}{4} \end{array}$

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