Lesson 9.5 : Solving Trigonometric Equations

New Concept

Solving trigonometric equations means finding the specific angle values that make an equation true. You'll learn to combine algebraic methods with trigonometric identities to solve linear, quadratic, and multi-angle equations, and apply these skills to real-world problems.

What’s next

Now, let's put this into practice. You'll start with interactive examples on linear equations, then tackle more complex types with a series of practice cards.

Property

Trigonometric equations can have an infinite number of solutions because the functions are periodic.
For the sine and cosine functions with a period of $2\pi$, if we find solutions in the interval $[0, 2\pi)$, all possible solutions can be found by adding $2\pi k$, where $k$ is an integer, to each solution.
To solve, use algebraic techniques: look for patterns, substitute the trigonometric expression with a variable like $x$ or $u$, solve the algebraic equation, substitute the trig expression back, and solve for the angle.

Examples

• Solve the equation $2\sin\theta + \sqrt{2} = 0$ on the interval $[0, 2\pi)$. First, isolate the sine: $2\sin\theta = -\sqrt{2}$, so $\sin\theta = -\frac{\sqrt{2}}{2}$. The angles in the given interval where this is true are $\theta = \frac{5\pi}{4}$ and $\theta = \frac{7\pi}{4}$.
• Find all possible exact solutions for $\cos\theta = -\frac{1}{2}$. On the unit circle, the solutions are $\theta = \frac{2\pi}{3}$ and $\theta = \frac{4\pi}{3}$. Since we need all solutions, we add multiples of $2\pi$, giving $\theta = \frac{2\pi}{3} + 2\pi k$ and $\theta = \frac{4\pi}{3} + 2\pi k$ for any integer $k$.
• Solve the equation $3\sin\theta - 2 = -\frac{1}{2}$ on $[0, 2\pi)$. Using algebra, we get $3\sin\theta = \frac{3}{2}$, so $\sin\theta = \frac{1}{2}$. The solutions on the interval are $\theta = \frac{\pi}{6}$ and $\theta = \frac{5\pi}{6}$.

Explanation

Because sine and cosine repeat their values every $2\pi$ radians, a single solution on the unit circle actually represents an entire family of solutions. We find the initial angles first, then add multiples of the period to capture all possibilities.

Property

When solving equations with a single trigonometric function, use algebraic techniques and the unit circle.
For equations with reciprocal functions, rewrite them in terms of the primary function (e.g., write $\operatorname{csc}\theta$ as $\frac{1}{\sin\theta}$).
For tangent functions, remember the period is $\pi$, not $2\pi$. The domain of tangent excludes odd integer multiples of $\frac{\pi}{2}$.

Examples

• Solve the equation $4\sin^2\theta - 1 = 0$ on $[0, 2\pi)$. We isolate the function: $\sin^2\theta = \frac{1}{4}$, so $\sin\theta = \pm\frac{1}{2}$. The solutions are $\theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$.
• Solve $\sec\theta = 2$ on $[0, 2\pi)$. First, rewrite as $\frac{1}{\cos\theta} = 2$, which means $\cos\theta = \frac{1}{2}$. The solutions on the interval are $\theta = \frac{\pi}{3}$ and $\theta = \frac{5\pi}{3}$.
• Solve the equation $\tan(\theta - \frac{\pi}{4}) = -1$ on $[0, 2\pi)$. The angle whose tangent is $-1$ is $\frac{3\pi}{4}$. So, $\theta - \frac{\pi}{4} = \frac{3\pi}{4}$, which gives $\theta = \pi$. Since the period of tangent is $\pi$, the next solution is $\theta = \pi + \pi = 2\pi$, which is outside the interval. The only solution is $\theta = \pi$.

Explanation

Think of these as regular algebra problems where your goal is to isolate the trig function. Once you have an expression like $\sin\theta = \text{value}$, you can use the unit circle to find the angles that make the statement true.

Property

When an equation involves an angle other than one of the special angles, a calculator is needed. Use the inverse trigonometric function to find the principal value solution.
For sine, the calculator returns an angle in Quadrant I or IV. The other angle is found using $\pi - \theta$.
For cosine, the calculator returns an angle in Quadrant I or II. The other angle is often found using $2\pi - \theta$ or by considering the reference angle in the appropriate quadrant.

Examples

• Use a calculator to solve $\sin\theta = 0.6$ in radians. The principal value is $\theta_1 = \sin^{-1}(0.6) \approx 0.6435$. Since sine is also positive in Quadrant II, the second solution is $\theta_2 = \pi - 0.6435 \approx 2.4981$.
• Use a calculator to solve $\cos\theta = -0.3$ in radians. The calculator gives a Quadrant II angle: $\theta_1 = \cos^{-1}(-0.3) \approx 1.8755$. Since cosine is also negative in Quadrant III, the second solution is $\theta_2 = 2\pi - 1.8755 \approx 4.4077$.
• Use a calculator to solve $\sec\theta = 5$ in radians. First, rewrite as $\cos\theta = \frac{1}{5} = 0.2$. The calculator gives $\theta_1 = \cos^{-1}(0.2) \approx 1.3694$. Since cosine is also positive in Quadrant IV, the second solution is $\theta_2 = 2\pi - 1.3694 \approx 4.9138$.

Explanation

When a value isn't on the unit circle, your calculator's inverse function is your best friend! But it only gives one answer. You must use your knowledge of reference angles and quadrants to find all other possible solutions.

Property

If a trigonometric equation contains a squared term, it may be quadratic in form.
To solve, first substitute the trigonometric function with a variable such as $x$ or $u$. This transforms the equation into a standard quadratic equation.
Solve for the variable using methods like factoring or the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Then, substitute the trigonometric function back in and solve for the angle.

Examples

• Solve $2\cos^2\theta - 7\cos\theta + 3 = 0$ on $[0, 2\pi)$. Factoring gives $(2\cos\theta - 1)(\cos\theta - 3) = 0$. This yields $\cos\theta = \frac{1}{2}$ or $\cos\theta = 3$. Since $\cos\theta$ cannot be 3, we solve $\cos\theta = \frac{1}{2}$ to get $\theta = \frac{\pi}{3}, \frac{5\pi}{3}$.
• Solve $\sin^2\theta + \sin\theta = 0$ on $[0, 2\pi)$. Factoring gives $\sin\theta(\sin\theta + 1) = 0$. This yields $\sin\theta = 0$ or $\sin\theta = -1$. The solutions are $\theta = 0, \pi$ and $\theta = \frac{3\pi}{2}$.
• Solve $\cos^2\theta + 2\cos\theta - 2 = 0$. Using the quadratic formula with $a=1, b=2, c=-2$, we get $\cos\theta = \frac{-2 \pm \sqrt{2^2 - 4(1)(-2)}}{2} = \frac{-2 \pm \sqrt{12}}{2} = -1 \pm \sqrt{3}$. Since $-1-\sqrt{3}$ is outside the range $[-1, 1]$, we only solve $\cos\theta = -1+\sqrt{3} \approx 0.732$. Using a calculator, $\theta \approx 0.75$ and $\theta \approx 2\pi - 0.75 = 5.53$.

Explanation

If an equation has a trig function that is squared, treat it like a quadratic! You can substitute the function with a variable like $u$ to make it a familiar algebra problem, solve for $u$, and then substitute back to find the angle.

Property

Fundamental identities can be used to simplify trigonometric equations, often by rewriting the equation in terms of a single trigonometric function.
Other useful identities include sum-and-difference formulas and double-angle formulas. The basic rules of algebra apply.

Examples

• Solve $2\cos^2\theta + 3\sin\theta - 3 = 0$ on $[0, 2\pi)$. Use the identity $\cos^2\theta = 1 - \sin^2\theta$ to get $2(1 - \sin^2\theta) + 3\sin\theta - 3 = 0$, which simplifies to $2\sin^2\theta - 3\sin\theta + 1 = 0$. Factoring gives $(2\sin\theta - 1)(\sin\theta - 1) = 0$. Solutions are $\theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{\pi}{2}$.
• Solve $\cos(2\theta) + \sin\theta = 0$ on $[0, 2\pi)$. Use the double-angle identity $\cos(2\theta) = 1 - 2\sin^2\theta$. The equation becomes $1 - 2\sin^2\theta + \sin\theta = 0$, or $2\sin^2\theta - \sin\theta - 1 = 0$. Factoring yields $(2\sin\theta + 1)(\sin\theta - 1) = 0$. Solutions are $\theta = \frac{7\pi}{6}, \frac{11\pi}{6}, \frac{\pi}{2}$.
• Solve $\cos(x)\cos(2x) + \sin(x)\sin(2x) = \frac{1}{2}$ on $[0, 2\pi)$. The left side is the difference formula for cosine, $\cos(x-2x) = \cos(-x) = \cos(x)$. So the equation is $\cos(x) = \frac{1}{2}$. The solutions are $x = \frac{\pi}{3}$ and $x = \frac{5\pi}{3}$.

Explanation

If you have a mix of sines and cosines, don't panic! Use fundamental identities to rewrite the equation in terms of a single trig function. This strategy simplifies the problem and makes it much easier to solve algebraically.

Property

For equations with a multiple angle, such as $\sin(nx)=c$, the period of the function is compressed. This means there may be more solutions in a given interval.
To solve, first find all solutions for the angle $nx$ by going around the unit circle $n$ times. Then, divide each resulting angle by $n$ to solve for $x$. Ensure the final solutions for $x$ are within the specified domain.

Examples

• Solve $\sin(2x) = \frac{\sqrt{3}}{2}$ on $[0, 2\pi)$. First, let $u=2x$. Then $\sin(u) = \frac{\sqrt{3}}{2}$ has solutions $u = \frac{\pi}{3}$ and $u = \frac{2\pi}{3}$. Since we need solutions for $x$ in $[0, 2\pi)$, we find solutions for $u$ in $[0, 4\pi)$: $u = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{7\pi}{3}, \frac{8\pi}{3}$. Dividing by 2 gives $x = \frac{\pi}{6}, \frac{\pi}{3}, \frac{7\pi}{6}, \frac{4\pi}{3}$.
• Solve $\cos(3x) = -1$ on $[0, 2\pi)$. Let $u=3x$. Then $\cos(u) = -1$. The solutions for $u$ in $[0, 6\pi)$ are $u = \pi, 3\pi, 5\pi$. Dividing by 3 gives the solutions for $x$: $x = \frac{\pi}{3}, \pi, \frac{5\pi}{3}$.
• Solve $\tan(2x) = 1$ on $[0, 2\pi)$. Let $u=2x$. Then $\tan(u)=1$. The solutions for $u$ in $[0, 4\pi)$ are $u = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}, \frac{13\pi}{4}$. Dividing by 2 gives $x = \frac{\pi}{8}, \frac{5\pi}{8}, \frac{9\pi}{8}, \frac{13\pi}{8}$.

Explanation

When you see an angle like $2x$ or $3x$, you're on the hunt for more solutions! First solve for the entire multiple angle ($nx$), finding all possible values, then divide by $n$ to get the solutions for $x$.

Property

Real-world problems involving right triangles can be modeled and solved using the Pythagorean Theorem, $a^2 + b^2 = c^2$, and the primary trigonometric functions. The relationships $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}$, $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}$, and $\tan\theta = \frac{\text{opposite}}{\text{adjacent}}$ are used to find unknown side lengths or angles.

Examples

• An 18-foot ladder leans against a building with its base 5 feet from the wall. What angle does the ladder make with the ground? The adjacent side is 5 ft and the hypotenuse is 18 ft. So, $\cos\theta = \frac{5}{18}$. The angle is $\theta = \cos^{-1}(\frac{5}{18}) \approx 73.9^\circ$.
• From a point 100 feet from the base of a tower, the angle of elevation to the top is $40^\circ$. How tall is the tower? Let the height be $h$. We have $\tan(40^\circ) = \frac{h}{100}$. So, the height is $h = 100 \tan(40^\circ) \approx 83.9$ feet.
• A kite is flying on a 150-meter string and the string makes an angle of $65^\circ$ with the horizontal ground. How high is the kite? Let the height be $h$. The string is the hypotenuse. We have $\sin(65^\circ) = \frac{h}{150}$. The height is $h = 150 \sin(65^\circ) \approx 135.9$ meters.

Explanation

Time to apply your skills! Draw a right triangle to represent the problem, label the sides and angles you know, and use the Pythagorean Theorem for lengths or SOHCAHTOA for finding missing angles or sides.