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Exercises Trigonometry

Example 12 from Algebra and Trigonometry, 9.5 Solving Trigonometric Equations

Solving a Trigonometric Equation Using Algebra

Solve exactly:

2 \sin^{2} θ + \sin θ = 0; 0 \leq θ < 2 π

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
This problem should appear familiar as it is similar to a quadratic. Let $\sin θ = x .$ The equation becomes $2 x^{2} + x = 0.$ We begin by factoring:

$x .$
Set each factor equal to zero.

$\begin{matrix} x & = & 0 \\ (2 x + 1) & = & 0 \\ x & = & - \frac{1}{2} \end{matrix}$
Then, substitute back into the equation the original expression $\sin θ$ for $x .$ Thus,
$\begin{matrix} \sin θ & = & 0 \\ θ & = & 0, π \\ \sin θ & = & - \frac{1}{2} \\ θ & = & \frac{7 π}{6}, \frac{11 π}{6} \end{matrix}$
The solutions within the domain $0 \leq θ < 2 π$ are $θ = 0, π, \frac{7 π}{6}, \frac{11 π}{6} .$

If we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero.

$\begin{matrix} 2 \sin^{2} θ + \sin θ & = & 0 \\ \sin θ (2 \sin θ + 1) & = & 0 \\ \sin θ & = & 0 \\ θ & = & 0, π \\ 2 \sin θ + 1 & = & 0 \\ 2 \sin θ & = & - 1 \\ \sin θ & = & - \frac{1}{2} \\ θ & = & \frac{7 π}{6}, \frac{11 π}{6} \end{matrix}$

How did it go?

PreviousSolving a Trigonometric Equation in Quadratic Form by Factoring Next Solving a Trigonometric Equation Quadratic in Form

From Algebra and Trigonometry (1st edition) by OpenStax (Rice University), section 9.5, licensed under CC BY 4.0. Original at openstax.org. We split the solution into steps; the words, numbers and figures are the book's. OpenStax doesn't endorse this site.