First slide

Trigonometric equations

Question

$I f θ \in [0, 5 π] a n d r \in R s u c h t h a t 2 \sin θ = r^{4} - 2 r^{2} + 3, t h e n t h e m a x i m u m n u m b e r o f v a l u e s o f t h e p a i r (r, θ) i s$

Moderate

A

8
B

10
C

6
D

4

Solution

$W e h a v e 2 \sin θ = r^{4} - 2 r^{2} + 3 = {(r^{2} - 1)}^{2} + 2 \geq 2 \Rightarrow 2 \sin θ \geq 2 \Rightarrow 2 \sin θ = 2 (∵ - 1 \leq \sin θ \leq 1) \Rightarrow \sin θ = 1$
$\Rightarrow θ = \frac{π}{2}, \frac{5 π}{2}, \frac{9 π}{2} (∵ θ \in [0, 5 π]) A l s o r^{2} - 1 = 0 \Rightarrow r = \pm 1 ∴ (r, θ) = (\pm 1, \frac{π}{2}), (\pm 1, \frac{5 π}{2}), (\pm 1, \frac{9 π}{2}) ∴ t h e r e a r e 6 p a i r s o f (r, θ)$

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