Solving Trigonometric Equations Exercise 2
Exercises
Solution: \(x=\dfrac{\pi}{4}, \dfrac{\pi}{2}, \dfrac{3\pi}{4},\dfrac{3\pi}{2}\)
Solution Method: A tempting first step here is to divide by a \(\cos{x}\) on both sides. But if we do that, we’ll be losing a solution to the equation! (Think about \(x^2=3x\). We can clearly see that \(x=0\) is a solution to the equation. But if we divide by x, we may think \(x=3\) is the only solution.)
So instead of dividing by \(\cos{x}\) as out first step, we’re going to move everything over to one side and set it equal to zero. \[\begin{aligned} 2\sin{x}\cos{x} &= \sqrt{2}\cos{x} \\ 2\sin{x}\cos{x} - \sqrt{2}\cos{x} &= 0 \end{aligned}\] Then I can factor out a \(\cos{x}\), \[\begin{aligned} 2\sin{x}\cos{x} - \sqrt{2}\cos{x} &= 0 \\ \cos{x}(2\sin{x} - \sqrt{2}) &= 0 \end{aligned}\] So I have two factors I can set equal to 0 and solve for x.
2 \[\begin{aligned} \cos{x} = 0 \end{aligned}\] Zero is a quadrantal value and thinking back to the unit circle, I know \(x=\frac{\pi}{2}, \frac{3\pi}{2}\) satisfy \(\cos{x}=0\). \[\begin{aligned} 2\sin{x}-\sqrt{2} &= 0 \\ 2\sin{x} = \sqrt{2} \\ \sin{x} = \frac{\sqrt{2}}{2} \end{aligned}\] The reference angle that produces sine equal to \(\frac{\sqrt{2}}{2}\) is \(\frac{\pi}{4}\). The sine value is also positive, so I’m in Q1 or Q2 (by ASTC). So the x values satisfying the equation are \(x=\frac{\pi}{4}, \frac{3\pi}{4}\).
So we have a total of four x-values satisfying \(2\sin{x}\cos{x} = \sqrt{2}\cos{x}\), \(x=\frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \frac{3\pi}{2}.\)
