Solved on Nov 07, 2023

Find the values of $x$ in the interval $[0, 2\pi]$ that satisfy the equation $2\csc^2 x - 8 = 0$ .

STEP 1

Assumptions1. The given equation is $\csc ^{} x-8=0$ . We are asked to find the solution set on the interval $[0, \pi]$
3. The cosecant function, $\csc(x)$ , is defined as $1/\sin(x)$ and is undefined for $x = n\pi$ where $n$ is an integer.

STEP 2

First, we need to solve the given equation for $\csc ^{2} x$ . We can do this by adding8 to both sides of the equation and then dividing by2.
$2 \csc ^{2} x =8$ $\csc ^{2} x = \frac{8}{2}$

STEP 3

implify the right side of the equation.
$\csc ^{2} x =$

STEP 4

Take the square root of both sides of the equation to solve for $\csc x$ . Remember that taking the square root of a number yields both a positive and negative solution.
$\csc x = \pm \sqrt{4}$

STEP 5

implify the right side of the equation.
$\csc x = \pm2$

STEP 6

Recall that $\csc x =1/\sin x$ . Therefore, we can rewrite the equation in terms of $\sin x$ .
$1/\sin x = \pm2$

STEP 7

olve for $\sin x$ by taking the reciprocal of both sides of the equation.
$\sin x = \pm \frac{1}{2}$

STEP 8

Now, we need to find the values of $x$ in the interval $[0,2 \pi]$ that satisfy the equation $\sin x = \pm \frac{1}{2}$ . Recall that $\sin x = \frac{1}{2}$ at $x = \pi/6$ and $x =5\pi/6$ , and $\sin x = -\frac{1}{2}$ at $x =7\pi/6$ and $x =11\pi/6$ .
So, the solution set on the interval $[0,2 \pi]$ is $\{ \pi/6,5\pi/6,7\pi/6,11\pi/6 \}$ .

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Find the values of $x$ in the interval $[0, 2\pi]$ that satisfy the equation $2\csc^2 x - 8 = 0$ .

STEP 1

Assumptions1. The given equation is $\csc ^{} x-8=0$ . We are asked to find the solution set on the interval $[0, \pi]$
3. The cosecant function, $\csc(x)$ , is defined as $1/\sin(x)$ and is undefined for $x = n\pi$ where $n$ is an integer.

STEP 2

First, we need to solve the given equation for $\csc ^{2} x$ . We can do this by adding8 to both sides of the equation and then dividing by2.
$2 \csc ^{2} x =8$ $\csc ^{2} x = \frac{8}{2}$

STEP 3

implify the right side of the equation.
$\csc ^{2} x =$

STEP 4

Take the square root of both sides of the equation to solve for $\csc x$ . Remember that taking the square root of a number yields both a positive and negative solution.
$\csc x = \pm \sqrt{4}$

STEP 5

implify the right side of the equation.
$\csc x = \pm2$

STEP 6

Recall that $\csc x =1/\sin x$ . Therefore, we can rewrite the equation in terms of $\sin x$ .
$1/\sin x = \pm2$

STEP 7

olve for $\sin x$ by taking the reciprocal of both sides of the equation.
$\sin x = \pm \frac{1}{2}$

STEP 8

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Find the values of xxx in the interval [0,2π][0, 2\pi][0,2π] that satisfy the equation 2csc⁡2x−8=02\csc^2 x - 8 = 02csc2x−8=0.

STEP 1

STEP 2

First, we need to solve the given equation for csc⁡2x\csc ^{2} xcsc2x. We can do this by adding8 to both sides of the equation and then dividing by2.2csc⁡2x=82 \csc ^{2} x =82csc2x=8csc⁡2x=82\csc ^{2} x = \frac{8}{2}csc2x=28​

STEP 3

implify the right side of the equation.csc⁡2x=\csc ^{2} x =csc2x=

STEP 4

Take the square root of both sides of the equation to solve for csc⁡x\csc xcscx. Remember that taking the square root of a number yields both a positive and negative solution.csc⁡x=±4\csc x = \pm \sqrt{4}cscx=±4​

STEP 5

implify the right side of the equation.csc⁡x=±2\csc x = \pm2cscx=±2

STEP 6

Recall that csc⁡x=1/sin⁡x\csc x =1/\sin xcscx=1/sinx. Therefore, we can rewrite the equation in terms of sin⁡x\sin xsinx.1/sin⁡x=±21/\sin x = \pm21/sinx=±2

STEP 7

olve for sin⁡x\sin xsinx by taking the reciprocal of both sides of the equation.sin⁡x=±12\sin x = \pm \frac{1}{2}sinx=±21​

STEP 8

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Find the values of $x$ in the interval $[0, 2\pi]$ that satisfy the equation $2\csc^2 x - 8 = 0$ .

First, we need to solve the given equation for $\csc ^{2} x$ . We can do this by adding8 to both sides of the equation and then dividing by2.
$2 \csc ^{2} x =8$ $\csc ^{2} x = \frac{8}{2}$

implify the right side of the equation.
$\csc ^{2} x =$

Take the square root of both sides of the equation to solve for $\csc x$ . Remember that taking the square root of a number yields both a positive and negative solution.
$\csc x = \pm \sqrt{4}$

implify the right side of the equation.
$\csc x = \pm2$

Recall that $\csc x =1/\sin x$ . Therefore, we can rewrite the equation in terms of $\sin x$ .
$1/\sin x = \pm2$

olve for $\sin x$ by taking the reciprocal of both sides of the equation.
$\sin x = \pm \frac{1}{2}$