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# Trigonometric Equations Using Factoring

## Factoring and the Quadratic Formula.

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Trigonometric Equations Using Factoring

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*}

### Trigonometric Equations Using Factoring

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.

#### Solving for Unknown Values

1. 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*}.

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*}

2. 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*}.

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*}.

3. 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*}.

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.

### Examples

#### Example 1

Earlier, you were asked 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*}

#### Example 2

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*}.

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*}

#### Example 3

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*}.

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*}

#### Example 4

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*}.

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*}.

### Review

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. 2cos(x)tan(x)tan(x)=0\begin{align*}2\cos(x)\tan(x)-\tan(x)=0\end{align*}
12. sin2(x)+sin(x)=2\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*}

To see the Review answers, open this PDF file and look for section 3.4.

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