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Exercises Trigonometry

Example 11 from Algebra and Trigonometry, 9.5 Solving Trigonometric Equations

Solving a Trigonometric Equation in Quadratic Form by Factoring

Solve the equation exactly: $2 \sin^{2} θ - 5 \sin θ + 3 = 0, 0 \leq θ \leq 2 π .$

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
Using grouping, this quadratic can be factored. Either make the real substitution, $\sin θ = u,$ or imagine it, as we factor:
$\begin{matrix} 2 \sin^{2} θ - 5 \sin θ + 3 & = & 0 \\ (2 \sin θ - 3) (\sin θ - 1) & = & 0 \end{matrix}$
Now set each factor equal to zero.

$\begin{matrix} 2 \sin θ - 3 & = & 0 \\ 2 \sin θ & = & 3 \\ \sin θ & = & \frac{3}{2} \\ \sin θ - 1 & = & 0 \\ \sin θ & = & 1 \end{matrix}$
Next solve for $θ : \sin θ \neq \frac{3}{2},$ as the range of the sine function is $[−1, 1] .$ However, $\sin θ = 1,$ giving the solution $θ = \frac{π}{2} .$

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Solve $\sin^{2} θ = 2 \cos θ + 2, 0 \leq θ \leq 2 π .$ [Hint: Make a substitution to express the equation only in terms of cosine.]

How did it go?

PreviousSolving a Trigonometric Equation in Quadratic Form Next Solving a Trigonometric Equation Using Algebra

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.