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tags:
Factoring Quadratic Sine Trigonometry Trigonometry Series

Solving Trigonometric Equations Exercise 1

Author: Hannah Solomon

The following problem is solved in this video. It is recommended that you try to solve the problem before watching the video. You can click "Reveal Answer" to see the answer to the problem.

Problem: Find all values \(0\leq x \leq 2\pi\) satisfying \(2\sin^2{x}+5\sin{x}-3=0.\)

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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