Solving Trigonometric Equations Exercise 1
Exercises
Solution: \(x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}\).
Solution Method: This equation is a "quadratic in sine". It’s in the form of \(ax^2+bx+c=0\), but the variable is \(\sin{x}\). Since the equation is in a quadratic form, I can factor it like I would the quadratic equation \(2x^2+5x-3=0.\) To factorize \(2x^2+5x-3=0\), we need to find two numbers whose product is -6 and sum to 5. So those numbers are going to be -1 and 6 \(( 6(-1)=-6, 6-1 = 5)\). Since the leading coefficient of the quadratic isn’t 1, we’re going to split the 5 into 6 and -1, and factor by grouping. \[\begin{aligned} 2x^2+5x-3&=0 \\ 2x^2 + 6x - x -3 &= 0 \\ 2x(x +3) - (x+3) &= 0 \\ (2x-1)(x+3) &= 0 \end{aligned}\] So then the factorization of the quadratic in sine is \[\begin{aligned} 2\sin^2{x}+5\sin{x}-3 &= 0 \\ (2\sin{x} - 1) (\sin{x}+3) &= 0 \end{aligned}\] We need to find the \(\sin{x}\) values, then the x-values, that satisfy the equation, so we’re going to take each factor and set it to zero.
\[\begin{aligned} 2\sin{x}-1 &= 0 \\ 2\sin{x} &= 1 \\ \sin{x} &= \frac{1}{2} \end{aligned}\] So now we think back to special angle values and the unit circle. Sine of what reference angle produces \(\frac{1}{2}\)? \(\frac{\pi}{6}\). The half is positive so our acronym ASTC reminds us that sine is positive in Q1 and Q2. So the x values (in \([0,2\pi]\)) that satisfy \(\sin{x}=\frac{1}{2}\) are \(x = \frac{\pi}{6}, \frac{5\pi}{6}\).
\[\begin{aligned} \sin{x}+3 &= 0 \\ \sin{x} &= -3 \\ \end{aligned}\]
Think back to the graph of a sine function. It’s range is [-1,1]. So is it ever possible for \(\sin{x} = -3\)? Nope. So this equation yields no solution.
So the values of \(x\) in \([0,2\pi]\) satisfying \(2\sin^2{x}+5\sin{x}-3=0\) are \(x = \frac{\pi}{6}, \frac{5\pi}{6}\).
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