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3.4: Trigonometric Equations Using Factoring

Difficulty Level: At Grade Created by: CK-12
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Practice Trigonometric Equations Using Factoring

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Solving trig equations is an important process in mathematics. Quite often you'll see powers of trigonometric functions and be asked to solve for the values of the variable which make the equation true. For example, suppose you were given the trig equation

2sinxcosx=cosx\begin{align*}2 \sin x \cos x = \cos x\end{align*}

Could you solve this equation? (You might be tempted to just divide both sides by cosx\begin{align*}\cos x\end{align*}, but that would be incorrect because you would lose some solutions.) Instead, you're going to have to use factoring. Read this Concept, and at its conclusion, you'll be ready to factor the above equation and solve it.

Guidance

You have no doubt had experience with factoring. You have probably factored equations when looking for the possible values of some variable, such as "x". It might interest you to find out that you can use the same factoring method for more than just a variable that is a number. You can factor trigonometric equations to find the possible values the function can take to satisfy an equation.

Algebraic skills like factoring and substitution that are used to solve various equations are very useful when solving trigonometric equations. As with algebraic expressions, one must be careful to avoid dividing by zero during these maneuvers.

Example A

Solve 2sin2x3sinx+1=0\begin{align*}2 \sin^2 x - 3 \sin x + 1 = 0\end{align*} for 0<x2π\begin{align*}0 < x \le 2 \pi\end{align*}.

Solution:

x2sin2x3sinx+1=0Factor this like a quadratic equation(2sinx1)(sinx1)=0     2sinx1=0or  sinx1=0   2sinx=1 sinx=1  sinx=12  x=π2=π6 and x=5π6\begin{align*}& \quad 2 \sin^2 x - 3 \sin x + 1 = 0 \quad \text{Factor this like a quadratic equation} \\ & (2 \sin x - 1)(\sin x - 1) = 0 \\ & \qquad \ \downarrow \qquad \qquad \ \ \ \searrow \\ & \ 2 \sin x - 1 = 0 \quad \text{or} \ \ \sin x - 1 = 0 \\ & \quad \ \ \ 2 \sin x = 1 \qquad \qquad \ \sin x = 1 \\ & \qquad \ \ \sin x = \frac{1}{2} \qquad \quad \qquad \ \ x = \frac{\pi}{2}\\ x & = \frac{\pi}{6} \ \text{and} \ x = \frac{5 \pi}{6}\end{align*}

Example B

Solve 2tanxsinx+2sinx=tanx+1\begin{align*}2 \tan x \sin x + 2 \sin x = \tan x + 1\end{align*} for all values of x\begin{align*}x\end{align*}.

Solution:

Pull out sinx\begin{align*}\sin x\end{align*}

There is a common factor of (tanx+1)\begin{align*}(\tan x + 1)\end{align*}

Think of the (tanx+1)\begin{align*}-(\tan x + 1)\end{align*} as (1)(tanx+1)\begin{align*}(-1)(\tan x + 1)\end{align*}, which is why there is a 1\begin{align*}-1\end{align*} behind the 2sinx\begin{align*}2 \sin x\end{align*}.

Example C

Solve 2sin2x+3sinx2=0\begin{align*}2 \sin^2 x + 3 \sin x - 2 = 0\end{align*} for all x,[0,π]\begin{align*}x, [0, \pi]\end{align*}.

Solution:

x2sin2x+3sinx2=0Factor like a quadratic(2sinx1)(sinx+2)=0  2sinx1=0sinx+2=0 sinx=12  sinx=2=π6 and x=5π6 There is no solution because the range of sinx is [1,1].\begin{align*}& \quad 2 \sin^2 x +3 \sin x - 2 = 0 \rightarrow \text{Factor like a quadratic} \\ & (2 \sin x -1)(\sin x + 2) = 0 \\ & \quad \ \ \swarrow \qquad \qquad \quad \searrow \\ & 2 \sin x - 1 = 0 \qquad \sin x + 2 = 0 \\ & \qquad \ \sin x = \frac{1}{2} \qquad \quad \ \ \sin x = -2 \\ x & = \frac{\pi}{6} \ \text{and} \ x = \frac{5 \pi}{6}\text{ There is no solution because the range of}\ \sin x\ \text{is}\ [-1, 1].\end{align*}

Some trigonometric equations have no solutions. This means that there is no replacement for the variable that will result in a true expression.

Vocabulary

Factoring: Factoring is a way to solve trigonometric equations by separating the equation into two terms which, when multiplied together, give the original expression. Since the product of the two factors is equal to zero, each of the factors can be equal to zero to make the original expression true. This leads to solutions for the original expression.

Guided Practice

1. Solve the trigonometric equation 4sinxcosx+2cosx2sinx1=0\begin{align*}4 \sin x \cos x + 2 \cos x-2 \sin x - 1 = 0\end{align*} such that 0x<2π\begin{align*}0 \le x < 2\pi\end{align*}.

2. Solve tan2x=3tanx\begin{align*}\tan^2 x = 3 \tan x\end{align*} for x\begin{align*}x\end{align*} over [0,π]\begin{align*}[0, \pi]\end{align*}.

3. Find all the solutions for the trigonometric equation 2sin2x43cosx4=0\begin{align*}2 \sin^2 \frac{x}{4}-3 \cos \frac{x}{4} = 0\end{align*} over the interval [0,2π)\begin{align*}[0, 2\pi)\end{align*}.

Solutions:

1. Use factoring by grouping.

2sinx+1=0or2cosx1=02sinx=12cosx=1  sinx=12cosx=12 x=7π6,11π6x=π3,5π3\begin{align*}& 2 \sin x + 1 = 0 \quad \text{or} \qquad 2 \cos x - 1 = 0 \\ & 2 \sin x = -1 \qquad \qquad \quad 2 \cos x = 1 \\ & \ \ \sin x = - \frac{1}{2} \qquad \qquad \quad \cos x = \frac{1}{2} \\ & \qquad \ x = \frac{7 \pi}{6}, \frac{11\pi}{6} \qquad \qquad \quad x = \frac{\pi}{3}, \frac{5\pi}{3}\end{align*}

2.

tan2xtan2x3tanxtanx(tanx3)tanxx=3tanx=0=0=0ortanx=3=0,π  x=1.25\begin{align*}\tan^2 x &= 3 \tan x \\ \tan^2 x - 3 \tan x &= 0 \\ \tan x (\tan x - 3) &= 0 \\ \tan x & = 0 \qquad \text{or} \qquad \tan x = 3 \\ x & = 0, \pi \qquad \qquad \quad \ \ x = 1.25\end{align*}

3.

2sin2x43cosx4=0\begin{align*}2 \sin^2 \frac{x}{4} - 3 \cos \frac{x}{4} = 0\end{align*}

2(1cos2x4)3cosx4=0 22cos2x43cosx4=0 2cos2x4+3cosx42=0(2cosx41)(cosx4+2)=02cosx41=0orcosx4+2=0  2cosx4=1  cosx4=2cosx4=12x4=π3or5π3x=4π3  or20π3\begin{align*}& \quad 2 \left (1 - \cos^2 \frac{x}{4} \right ) - 3 \cos \frac{x}{4} = 0 \\ & \qquad \ 2 - 2 \cos^2 \frac{x}{4} - 3 \cos \frac{x}{4} = 0 \\ & \qquad \ 2 \cos^2 \frac{x}{4} + 3 \cos \frac{x}{4} - 2 = 0 \\ & \left (2 \cos \frac{x}{4} - 1 \right ) \left (\cos \frac{x}{4} + 2 \right ) = 0 \\ & \qquad \swarrow \qquad \qquad \qquad \searrow\\ & 2 \cos \frac{x}{4} - 1 = 0 \quad \text{or} \quad \cos \frac{x}{4} + 2 = 0 \\ & \quad \ \ 2 \cos \frac{x}{4} = 1 \qquad \qquad \ \ \cos \frac{x}{4} = -2 \\ & \qquad \cos \frac{x}{4} = \frac{1}{2} \\ & \frac{x}{4} = \frac{\pi}{3} \quad \text{or} \quad \frac{5\pi}{3} \\ & x = \frac{4 \pi}{3} \ \ \text{or} \quad \frac{20\pi}{3}\end{align*}

20π3\begin{align*}\frac{20 \pi}{3}\end{align*} is eliminated as a solution because it is outside of the range and cosx4=2\begin{align*}\cos \frac{x}{4} = -2\end{align*} will not generate any solutions because 2\begin{align*}-2\end{align*} is outside of the range of cosine. Therefore, the only solution is 4π3\begin{align*}\frac{4 \pi}{3}\end{align*}.

Concept Problem Solution

The equation you were given is

2sinxcosx=cosx\begin{align*}2 \sin x \cos x = \cos x\end{align*}

To solve this:

2sinxcosx=cosx\begin{align*}2 \sin x \cos x = \cos x\end{align*}

Subtract cosx\begin{align*}\cos x\end{align*} from both sides and factor it out of the equation:

2sinxcosxcosx=0cosx(2sinx1)=0\begin{align*} 2 \sin x \cos x - \cos x = 0\\ \cos x (2 \sin x - 1) = 0\\ \end{align*}

Now set each factor equal to zero and solve. The first is cosx\begin{align*}\cos x\end{align*}:

cosx=0x=π2,3π2\begin{align*} \cos x = 0\\ x = \frac{\pi}{2}, \frac{3\pi}{2}\\ \end{align*}

And now for the other term:

2sinx1=0sinx=12x=π6,5π6\begin{align*} 2 \sin x - 1 = 0\\ \sin x = \frac{1}{2}\\ x = \frac{\pi}{6}, \frac{5\pi}{6}\\ \end{align*}

Practice

Solve each equation for x\begin{align*}x\end{align*} over the interval [0,2π)\begin{align*}[0,2\pi)\end{align*}.

1. cos2(x)+2cos(x)+1=0\begin{align*}\cos^2(x)+2\cos(x)+1=0\end{align*}
2. 12sin(x)+sin2(x)=0\begin{align*}1-2\sin(x)+\sin^2(x)=0\end{align*}
3. 2cos(x)sin(x)cos(x)=0\begin{align*}2\cos(x)\sin(x)-\cos(x)=0\end{align*}
4. sin(x)tan2(x)sin(x)=0\begin{align*}\sin(x)\tan^2(x)-\sin(x)=0\end{align*}
5. sec2(x)=4\begin{align*}\sec^2(x)=4\end{align*}
6. sin2(x)2sin(x)=0\begin{align*}\sin^2(x)-2\sin(x)=0\end{align*}
7. 3sin(x)=2cos2(x)\begin{align*}3\sin(x)=2\cos^2(x)\end{align*}
8. 2sin2(x)+3sin(x)=2\begin{align*}2\sin^2(x)+3\sin(x)=2\end{align*}
9. tan(x)sin2(x)=tan(x)\begin{align*}\tan(x)\sin^2(x)=\tan(x)\end{align*}
10. 2sin2(x)+sin(x)=1\begin{align*}2\sin^2(x)+\sin(x)=1\end{align*}
11. \begin{align*}2\cos(x)\tan(x)-\tan(x)=0\end{align*}
12. \begin{align*}\sin^2(x)+\sin(x)=2\end{align*}
13. \begin{align*}\tan(x)(2\cos^2(x)+3\cos(x)-2)=0\end{align*}
14. \begin{align*}\sin^2(x)+1=2\sin(x)\end{align*}
15. \begin{align*}2\cos^2(x)-3\cos(x)=2\end{align*}

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Vocabulary Language: English

Factoring

Factoring is the process of dividing a number or expression into a product of smaller numbers or expressions.

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