First slide

Trigonometric equations

Question

$If 4 x^{2} + 2 \cos^{2} θ = 4 x - \sin^{2} θ then the maximum number of ordered pairs (x, θ) in [0, 2 π] is$

Moderate

A

0
B

1
C

2
D

infinite

Solution

$\begin{array}{l} 4 x^{2} - 4 x + 1 + \cos^{2} θ = 0 \\ \Rightarrow (2 x - 1)^{2} + \cos^{2} θ = 0 \\ \Rightarrow 2 x - 1 = 0 and \cos θ = 0 \\ \Rightarrow x = \frac{1}{2} and θ = \frac{π}{2}, \frac{3 π}{2} \\ \Rightarrow (\frac{1}{2}, \frac{π}{2}); (\frac{1}{2}, \frac{3 π}{2}) \end{array}$
Thus, 2 ordered pairs are possible.

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