Question #aa149

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Question #aa149

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Answer 1 · 2017-02-18T16:44:33

$x = 0, \frac{π}{4}, \frac{3 π}{4}, π, \frac{5 π}{4}, \frac{7 π}{4}$ $x = 0, pi/4, (3pi)/4, pi, (5pi)/4, (7pi)/4$

Explanation:

First, set the equation equal to zero. Do this by subtracting the $sin$ $sin$ on the right side.

$sin x {tan}^{2} x - sin x = 0$ $sinxtan^2x - sinx = 0$

Now, noticing that each term has at least one $sin$ $sin$ in it, factor out one $sin$ $sin$ .

$sin x ({tan}^{2} x - 1) = 0$ $color(blue)(sinx)color(red)((tan^2x - 1)) = 0$

We have made it much easier to solve! Set each factor (whatever is being multiplied) equal to zero and solve.

$sin x = 0$ $color(blue)(sin x) =0$ ...and... ${tan}^{2} x - 1 = 0$ $color(red)(tan^2x-1) = 0$

Let's being with the equation on the left.

$sin x = 0$ $color(blue)(sin x) = 0$

Think back to the location on the unit circle where $sin$ $sin$ equals zero. $sin$ $sin$ is refers to the $y$ $y$ value, so at what unit circle point does the $y$ $y$ value of the coordinate rest at the $x$ $x$ axis? I've included a unit circle below...

![GradeAMathHelp.com](useruploads.socratic.org)

This happens when the radian value is $0$ $0$ or $π$ $pi$ .

$x = 0$ $color(blue)(x) = 0$

$x = π$ $color(blue)(x) = pi$

We have found two solutions for $x$ $x$ so far. Now we need to go back and do the same with the other function.

${tan}^{2} x - 1 = 0$ $color(red)(tan^2x-1) = 0$

To solve, begin by isolating the ${tan}^{2} x$ $tan^2x$ .

${tan}^{2} x = 1$ $color(red)(tan^2x) = 1$

Next we have to take the square root of both sides.

$\sqrt{{tan}^{2} x} = \sqrt{1}$ $color(red)(sqrt(tan^2x)) = sqrt(1)$

Since the square root of something squared is itself and the square root of $1$ $1$ is itself, let's rewrite this. Also note that the square root of $1$ $1$ is negative and positive $1$ $1$ .

$tan x = 1$ $color(red)(tanx) = 1$ ...and... $tan x = - 1$ $color(red)(tanx) = -1$

At this point, we have to think back again to the unit circle. Since $tan$ $tan$ refers to the $y$ $y$ value divided by the $x$ $x$ value, or $sin$ $sin$ over $cos$ $cos$ , think to any points or values where the quotient is $1$ $1$ . This typically true when the $x$ $x$ and $y$ $y$ value of the coordinates are identical. I've added a unit circle for reference again.

![GradeAMathHelp.com](useruploads.socratic.org)

This is true when the radian value is $\frac{π}{4}$ $pi/4$ and $\frac{5 π}{4}$ $(5pi)/4$ , since the $x$ $x$ and $y$ $y$ values are the same in each of their respective coordinates. This is also true when the radian value is $\frac{3 π}{4}$ $(3pi)/4$ and $\frac{7 π}{4}$ $(7pi)/4$ , because their $sin$ $sin$ over $cos$ $cos$ value is $- 1$ $-1$ .

$x = \frac{π}{4}$ $color(red)(x) = pi/4$

$x = \frac{3 π}{4}$ $color(red)(x) = (3pi)/4$

$x = \frac{5 π}{4}$ $color(red)(x) = (5pi)/4$

$x = \frac{7 π}{4}$ $color(red)(x) = (7pi)/4$

These are all of the solutions we found. Let's review our steps:

1.) Get all terms to one side and set the equation equal to zero.

2). Factor (doesn't work in every equation, but was useful here).

3). Set each factor equal to zero and solve using the unit circle.

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Question #aa149

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