Solving Trigonometric Equations Part II Lesson

Unit Circle

The unit circle will be given here for reference.

Solving Trigonometric Equations Using Factoring

In the last lesson, we learned the basics of how to solve a trigonometric equation using linear methods. In some cases, we may need to factor in order to solve a trigonometric equation. Let's look at a few examples.
Example #1: Solve each equation over the interval [0°, 360°) and for all solutions. $$3\text{cos}\hspace{.1em}θ + 2\text{cos}^2 θ=-1$$ Add 1 to each side of the equation and rearrange: $$2\text{cos}^2 θ + 3\text{cos}\hspace{.1em}θ + 1=0$$ Factor:
It sometimes helps to think of cos θ as u: $$2u^2 + 3u + 1=0$$ $$(2u + 1)(u + 1)=0$$ $$2\text{cos}^2 θ + 3\text{cos}\hspace{.1em}θ + 1=0$$ $$(2\text{cos}\hspace{.1em}θ + 1)(\text{cos}\hspace{.1em}θ + 1)=0$$ Use the Zero-Product Property: $$2\text{cos}\hspace{.1em}θ + 1=0$$ $$\text{or}$$ $$\text{cos}\hspace{.1em}θ + 1=0$$ Let's start with the top equation: $$2\text{cos}\hspace{.1em}θ + 1=0$$ Subtract 1 away from each side: $$2\text{cos}\hspace{.1em}θ=-1$$ Divide each side by 2: $$\text{cos}\hspace{.1em}θ=-\frac{1}{2}$$ Again, if we reference our unit circle above, we see solutions of 120° and 240°.
If we wanted to use a calculator: $$\text{cos}^{-1}\left(-\frac{1}{2}\right)=120°$$ Note, the reference angle is 180° - 120° = 60°.
What angle in quadrant III has a reference angle of 60°? $$180° + 60°=240°$$ Now, let's work on our bottom equation. $$\text{cos}\hspace{.1em}θ + 1=0$$ Subtract 1 away from each side: $$\text{cos}\hspace{.1em}θ=-1$$ From the unit circle above, we see a single solution of 180°.
Again, we can use our calculator: $$\text{cos}^{-1}(-1)=180°$$ We can put all of our solutions together and obtain: $$\{120°, 180°, 240°\}$$ or using radians: $$\left\{\frac{2π}{3}, π, \frac{4π}{3}\right\}$$ For all solutions, we simply add 360°n to each solution when using degrees or $2π$n when using radians. $$\{120° + 360°n, 180° + 360°n, 240° + 360°n\}$$ or using radians: $$\left\{\frac{2π}{3}+ 2πn, π + 2πn, \frac{4π}{3}+ 2πn\right\}$$ Let's look at another example.
Example #2: Solve each equation over the interval $[0, 2π)$ and for all solutions. $$4=\text{sin}\hspace{.1em}β + 2\text{sin}^2 β + 3$$ Subract 4 away from each side: $$0=\text{sin}\hspace{.1em}β + 2\text{sin}^2 β - 1$$ Switch Sides: $$\text{sin}\hspace{.1em}β + 2\text{sin}^2 β - 1=0$$ Rearrange in the form: $ax^2 + bx + c=0$ $$2\text{sin}^2 β + \text{sin}\hspace{.1em}β - 1=0$$ To factor, we can replace sin β with u. This isn't necessary but it is helpful for some students: $$2u^2 + u - 1=0$$ $$(2u - 1)(u + 1)=0$$ Replace u with sin β: $$(2u - 1)(u + 1)=0$$ $$(2\text{sin}\hspace{.1em}β - 1)(\text{sin}\hspace{.1em}β + 1)=0$$ Solve Using the Zero Product Property: $$2\text{sin}\hspace{.1em}β - 1=0$$ $$\text{or}$$ $$\text{sin}\hspace{.1em}β + 1=0$$ Let's begin with the top equation: $$2\text{sin}\hspace{.1em}β - 1=0$$ Add 1 to each side: $$2\text{sin}\hspace{.1em}β=1$$ Divide each side by 2: $$\text{sin}\hspace{.1em}β=\frac{1}{2}$$ From the unit circle, we see our solutions for β are 30° or 150°. In terms of radians, we get: $$\frac{π}{6}, \frac{5π}{6}$$ Using a calculator, we would find that: $$\text{sin}^{-1}\left(\frac{1}{2}\right)=30°$$ Again, since sine is positive in quadrants I and II, we can find the angle in quadrant II with a reference angle of 30°. This will be our other solution. $$180° - 30°=150°$$ Our two solutions for β are 30° and 150°.
Let's look at our other equation: $$\text{sin}\hspace{.1em}β + 1=0$$ Subtract 1 from each side: $$\text{sin}\hspace{.1em}β=-1$$ From the unit circle, we see our solution for β is 270° or in terms of radians: $$\frac{3π}{2}$$ Again, from our calculator: $$\text{sin}^{-1}(-1)=-90°$$ Since we are working from 0 to $2π$, we can add $2π$ or 360° to this to obtain our 270°. We can put all of our solutions together and obtain: $$\{30°, 150°, 270°\}$$ or using radians: $$\left\{\frac{π}{6}, \frac{5π}{6}, \frac{3π}{2}\right\}$$ For all solutions with the sine function, we simply add 360°n to each solution when using degrees or $2π$n when using radians. In this case, notice how going from 30° to 150° to 270° is a jump of 120° each time. This allows us to write our general solution as: $$\{30° + 120°n\}$$ $$\left\{\frac{π}{6}+ \frac{2π}{3}n\right\}$$ Let's look at another example.
Example #3: Solve each equation over the interval $[0, 2π)$ and for all solutions. $$\text{tan}^{2}β + \text{tan}β - 2=0$$ Let's factor the left side: $$(\text{tan}(β) - 1)(\text{tan}(β) + 2)=0$$ Solve Using the Zero Product Property: $$\text{tan}(β) - 1=0$$ $$\text{or}$$ $$\text{tan}(β) + 2=0$$ Let's begin with the top equation. $$\text{tan}(β) - 1=0$$ Add 1 to both sides: $$\text{tan}(β)=1$$ $$\text{tan}^{-1}(1)=\frac{\pi}{4}$$ One solution is $\frac{\pi}{4}$ or 45°. Tangent is positive in quadrants I and III, we can find the angle in quadrant III with a reference angle of $\frac{\pi}{4}$ or 45°. This will be our other solution. $$\frac{4π + \pi}{4}=\frac{5\pi}{4}$$ $$45° + 180°=225°$$ Let's look at our other equation: $$\text{tan}(β) + 2=0$$ Subtract 2 away from each side: $$\text{tan}(β)=-2$$ To keep things simple, let's use the inverse tangent function with positive 2 as the argument. Note: You can also use (-2) as the argument here but you will obtain an angle in quadrant IV rotating clockwise. In this case, you need to add $\pi$ to obtain one solution and $2\pi$ to obtain the other. $$\text{tan}^{-1}(2) ≈ 1.107$$ $$\text{tan}^{-1}(2) ≈ 63.43°$$ The above gives us a reference angle (radians on the top, degrees on the bottom) for our solutions in quadrants II and IV, where tangent is negative.
Quadrant II: $$β=180° - 63.43°=116.57°$$ $$β=\pi - 1.107=2.03$$ Note: the above answers are both approximations. The exact value would be given as: $$β=180° - \text{tan}^{-1}(2)$$ $$β=\pi - \text{tan}^{-1}(2)$$ Quadrant IV: $$β=360° - 63.43°=296.57°$$ $$β=2\pi - 1.107=5.18$$ Note: the above answers are both approximations. The exact value would be given as: $$β=360° - \text{tan}^{-1}(2)$$ $$β=2\pi - \text{tan}^{-1}(2)$$ We can put all of our solutions together and obtain: $$\left\{45°, 116.57°, 225°, 296.57°\right\}$$ or using radians: $$\left\{\frac{π}{4}, 2.03, \frac{5π}{4}, 5.18\right\}$$ For all solutions, the period of tangent is $\pi$ or 180°. $$\left\{\frac{π}{4}+ πn, 2.03 + πn\right\}$$ $$\left\{45° + 180°n, 116.57 + 180°n\right\}$$

Problems When Dividing Both Sides by a Trigonometric Function

When solving a trigonometric equation, problems can arise when we divide both sides by a trigonometric function. Let's take a look at a simple example.
Example #4: Solve each equation over the interval $[0, 2π)$ and for all solutions. $$\text{cos}β \cdot \text{cot}β=\text{cos}β$$ Let's begin by using factoring to solve the equation.
Subtract cos β away from each side: $$\text{cos}β \cdot \text{cot}β - \text{cos}β=0$$ Factor out cos β: $$\text{cos}β(\text{cot}β - 1)=0$$ Solve Using the Zero Product Property: $$\text{cos}β=0$$ $$\text{or}$$ $$\text{cot}β - 1=0$$ Let's begin with the top equation: $$\text{cos}β=0$$ From the unit circle, we see our solutions for β are 90° or 270°. In terms of radians, we get: $$\frac{\pi}{2}, \frac{3\pi}{2}$$ Let's look at our other equation: $$\text{cot}β - 1=0$$ Add 1 to each side: $$\text{cot}β=1$$ Use the reciprocal identity: $$\text{tan}β=\frac{1}{\text{cot}β}$$ $$\text{tan}β=1$$ Use the inverse tangent function. $$\text{tan}^{-1}(1)=\frac{\pi}{4}$$ We know the period of tangent is $\pi$ or 180°. This means we would also have a solution in quadrant III. $$\pi + \frac{\pi}{4}=\frac{4\pi + \pi}{4}=\frac{5\pi}{4}$$ Let's put all of our solutions together and obtain: $$\left\{\frac{\pi}{4}, \frac{\pi}{2}, \frac{5\pi}{4}, \frac{3\pi}{2}\right\}$$ or using degrees: $$\left\{45°, 90°, 225°, 270°\right\}$$ For all solutions: $$\left\{\frac{\pi}{4}+ πn, \frac{\pi}{2}+ πn\right\}$$ or using degrees: $$\left\{45° + 180°n, 90° + 180°n\right\}$$ In this case, it may be helpful to look at the solutions using a basic sketch. Let's try to solve the problem using a different approach. Let's suppose we decided to divide both sides of the equation by cos β. $$\text{cos}β \cdot \text{cot}β=\text{cos}β$$ $$\frac{\text{cos}β \cdot \text{cot}β}{\text{cos}β}=\frac{\text{cos}β}{\text{cos}β}$$ $$\text{cot}β=1$$ Here, we would only get that β is 45° or 225° ($\frac{\pi}{4}$ or $\frac{5\pi}{4}$). What happened to the solutions of 90° and 270° ($\frac{\pi}{2}$ and $\frac{3\pi}{2}$) we found earlier? We are missing the solutions from the original equation that make the divisor zero. $$\text{cos}\frac{\pi}{2}=0$$ $$\text{cos}\frac{3\pi}{2}=0$$ For this reason, it is recommended that we don't divide both sides by a trigonometric function such as cos β.

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