Solved on Oct 31, 2023

Prove $2 \cot 2x + \tan x \equiv \cot x$ for $x \neq n\pi/2, n \in \mathbb{Z}$ . Then solve $6 \cot 2x + 3 \tan x = \csc^2 x - 2$ for $-\pi \leq x < \pi$ , giving answers to 3 decimal places. (No graphical/numerical solutions).

STEP 1

Assumptions1. The variable $x$ is not equal to $\frac{n\pi}{}$ , where $n$ is an integer. . The trigonometric identities are valid for all values of $x$ .

STEP 2

We start by proving the first equation. We can use the identity $\cot(2x) = \frac{1 - \tan^2(x)}{2\tan(x)}$ to rewrite the left-hand side of the equation.
$2\cot(2x) + \tan(x) =2\left(\frac{1 - \tan^2(x)}{2\tan(x)}\right) + \tan(x)$

STEP 3

implify the equation.
$2\cot(2x) + \tan(x) = \frac{1 - \tan^2(x)}{\tan(x)} + \tan(x)$

STEP 4

Multiply through by $\tan(x)$ to clear the fraction.
$\tan(x)\left(2\cot(2x) + \tan(x)\right) =1 - \tan^2(x) + \tan^2(x)$

STEP 5

implify the equation.
$2\cot(2x)\tan(x) + \tan^2(x) =1$

STEP 6

Substitute $\cot(2x)\tan(x)$ with $\frac{1}{2}$ (since $\cot(x)\tan(x) =1$ ) and simplify the equation.
$2\left(\frac{1}{2}\right) + \tan^2(x) =1$

STEP 7

implify the equation.
$1 + \tan^2(x) =1$

STEP 8

Subtract1 from both sides of the equation.
$\tan^2(x) =0$

STEP 9

Take the square root of both sides of the equation.
$\tan(x) =$

STEP 10

Substitute $\tan(x)$ with $\cot(\frac{\pi}{2} - x)$ .
$\cot(\frac{\pi}{2} - x) =0$

STEP 11

This is equivalent to $\cot(x)$ , so the first equation is proven.

STEP 12

Now we solve the second equation. Substitute $2\cot(2x) + \tan(x)$ with $\cot(x)$ in the second equation.
$6\cot(x) +\tan(x) = \csc^2(x) -2$

STEP 13

Multiply through by $\sin^2(x)$ to clear the fraction.
$6\cot(x)\sin^2(x) +3\tan(x)\sin^2(x) = -2\sin^2(x)$

STEP 14

Substitute $\cot(x)\sin^2(x)$ with $\cos^2(x)$ and $\tan(x)\sin^2(x)$ with $\sin^2(x)$ .
$6\cos^2(x) +3\sin^2(x) = -2\sin^2(x)$

STEP 15

Rearrange the equation to find the value of $\sin^2(x)$ .
$5\sin^2(x) = -\cos^2(x)$

STEP 16

Substitute $\cos^2(x)$ with $- \sin^2(x)$ .
$5\sin^2(x) = -6( - \sin^2(x))$

STEP 17

implify the equation.
$5\sin^2(x) = -6 +6\sin^2(x)$

STEP 18

Rearrange the equation to find the value of $\sin^2(x)$ .
$\sin^2(x) = \frac{5}{11}$

STEP 19

Take the square root of both sides of the equation.
$\sin(x) = \pm\sqrt{\frac{5}{11}}$

STEP 20

Calculate the value of $x$ for $-\pi \leq x < \pi$ .
$x = \sin^{-}\left(\pm\sqrt{\frac{5}{11}}\right)$

STEP 21

Calculate the values of $x$ to3 decimal places.
$x \approx \pm0.857, \pm.285$ The solutions are $x \approx0.857, -0.857,.285, -.285$ .

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STEP 1

Assumptions1. The variable xxx is not equal to nπ\frac{n\pi}{}nπ​, where nnn is an integer. . The trigonometric identities are valid for all values of xxx.

STEP 2

STEP 3

implify the equation.2cot⁡(2x)+tan⁡(x)=1−tan⁡2(x)tan⁡(x)+tan⁡(x)2\cot(2x) + \tan(x) = \frac{1 - \tan^2(x)}{\tan(x)} + \tan(x)2cot(2x)+tan(x)=tan(x)1−tan2(x)​+tan(x)

STEP 4

Multiply through by tan⁡(x)\tan(x)tan(x) to clear the fraction.tan⁡(x)(2cot⁡(2x)+tan⁡(x))=1−tan⁡2(x)+tan⁡2(x)\tan(x)\left(2\cot(2x) + \tan(x)\right) =1 - \tan^2(x) + \tan^2(x)tan(x)(2cot(2x)+tan(x))=1−tan2(x)+tan2(x)

STEP 5

implify the equation.2cot⁡(2x)tan⁡(x)+tan⁡2(x)=12\cot(2x)\tan(x) + \tan^2(x) =12cot(2x)tan(x)+tan2(x)=1

STEP 6

Substitute cot⁡(2x)tan⁡(x)\cot(2x)\tan(x)cot(2x)tan(x) with 12\frac{1}{2}21​ (since cot⁡(x)tan⁡(x)=1\cot(x)\tan(x) =1cot(x)tan(x)=1) and simplify the equation.2(12)+tan⁡2(x)=12\left(\frac{1}{2}\right) + \tan^2(x) =12(21​)+tan2(x)=1

STEP 7

implify the equation.1+tan⁡2(x)=11 + \tan^2(x) =11+tan2(x)=1

STEP 8

Subtract1 from both sides of the equation.tan⁡2(x)=0\tan^2(x) =0tan2(x)=0

STEP 9

Take the square root of both sides of the equation.tan⁡(x)=\tan(x) =tan(x)=

STEP 10

Substitute tan⁡(x)\tan(x)tan(x) with cot⁡(π2−x)\cot(\frac{\pi}{2} - x)cot(2π​−x).cot⁡(π2−x)=0\cot(\frac{\pi}{2} - x) =0cot(2π​−x)=0

STEP 11

This is equivalent to cot⁡(x)\cot(x)cot(x), so the first equation is proven.

STEP 12

Now we solve the second equation. Substitute 2cot⁡(2x)+tan⁡(x)2\cot(2x) + \tan(x)2cot(2x)+tan(x) with cot⁡(x)\cot(x)cot(x) in the second equation.6cot⁡(x)+tan⁡(x)=csc⁡2(x)−26\cot(x) +\tan(x) = \csc^2(x) -26cot(x)+tan(x)=csc2(x)−2

STEP 13

Multiply through by sin⁡2(x)\sin^2(x)sin2(x) to clear the fraction.6cot⁡(x)sin⁡2(x)+3tan⁡(x)sin⁡2(x)=−2sin⁡2(x)6\cot(x)\sin^2(x) +3\tan(x)\sin^2(x) = -2\sin^2(x)6cot(x)sin2(x)+3tan(x)sin2(x)=−2sin2(x)

STEP 14

Substitute cot⁡(x)sin⁡2(x)\cot(x)\sin^2(x)cot(x)sin2(x) with cos⁡2(x)\cos^2(x)cos2(x) and tan⁡(x)sin⁡2(x)\tan(x)\sin^2(x)tan(x)sin2(x) with sin⁡2(x)\sin^2(x)sin2(x).6cos⁡2(x)+3sin⁡2(x)=−2sin⁡2(x)6\cos^2(x) +3\sin^2(x) = -2\sin^2(x)6cos2(x)+3sin2(x)=−2sin2(x)

STEP 15

Rearrange the equation to find the value of sin⁡2(x)\sin^2(x)sin2(x).5sin⁡2(x)=−cos⁡2(x)5\sin^2(x) = -\cos^2(x)5sin2(x)=−cos2(x)

STEP 16

Substitute cos⁡2(x)\cos^2(x)cos2(x) with −sin⁡2(x) - \sin^2(x)−sin2(x).5sin⁡2(x)=−6(−sin⁡2(x))5\sin^2(x) = -6( - \sin^2(x))5sin2(x)=−6(−sin2(x))

STEP 17

implify the equation.5sin⁡2(x)=−6+6sin⁡2(x)5\sin^2(x) = -6 +6\sin^2(x)5sin2(x)=−6+6sin2(x)

STEP 18

Rearrange the equation to find the value of sin⁡2(x)\sin^2(x)sin2(x).sin⁡2(x)=511\sin^2(x) = \frac{5}{11}sin2(x)=115​

STEP 19

Take the square root of both sides of the equation.sin⁡(x)=±511\sin(x) = \pm\sqrt{\frac{5}{11}}sin(x)=±115​​

STEP 20

Calculate the value of xxx for −π≤x<π-\pi \leq x < \pi−π≤x<π.x=sin⁡−(±511)x = \sin^{-}\left(\pm\sqrt{\frac{5}{11}}\right)x=sin−(±115​​)

STEP 21

Calculate the values of xxx to3 decimal places.x≈±0.857,±.285x \approx \pm0.857, \pm.285x≈±0.857,±.285The solutions are x≈0.857,−0.857,.285,−.285x \approx0.857, -0.857,.285, -.285x≈0.857,−0.857,.285,−.285.

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Assumptions1. The variable $x$ is not equal to $\frac{n\pi}{}$ , where $n$ is an integer. . The trigonometric identities are valid for all values of $x$ .

implify the equation.
$2\cot(2x) + \tan(x) = \frac{1 - \tan^2(x)}{\tan(x)} + \tan(x)$

Multiply through by $\tan(x)$ to clear the fraction.
$\tan(x)\left(2\cot(2x) + \tan(x)\right) =1 - \tan^2(x) + \tan^2(x)$

implify the equation.
$2\cot(2x)\tan(x) + \tan^2(x) =1$

Substitute $\cot(2x)\tan(x)$ with $\frac{1}{2}$ (since $\cot(x)\tan(x) =1$ ) and simplify the equation.
$2\left(\frac{1}{2}\right) + \tan^2(x) =1$

implify the equation.
$1 + \tan^2(x) =1$

Subtract1 from both sides of the equation.
$\tan^2(x) =0$

Take the square root of both sides of the equation.
$\tan(x) =$

Substitute $\tan(x)$ with $\cot(\frac{\pi}{2} - x)$ .
$\cot(\frac{\pi}{2} - x) =0$

This is equivalent to $\cot(x)$ , so the first equation is proven.

Now we solve the second equation. Substitute $2\cot(2x) + \tan(x)$ with $\cot(x)$ in the second equation.
$6\cot(x) +\tan(x) = \csc^2(x) -2$

Multiply through by $\sin^2(x)$ to clear the fraction.
$6\cot(x)\sin^2(x) +3\tan(x)\sin^2(x) = -2\sin^2(x)$

Substitute $\cot(x)\sin^2(x)$ with $\cos^2(x)$ and $\tan(x)\sin^2(x)$ with $\sin^2(x)$ .
$6\cos^2(x) +3\sin^2(x) = -2\sin^2(x)$

Rearrange the equation to find the value of $\sin^2(x)$ .
$5\sin^2(x) = -\cos^2(x)$

Substitute $\cos^2(x)$ with $- \sin^2(x)$ .
$5\sin^2(x) = -6( - \sin^2(x))$

implify the equation.
$5\sin^2(x) = -6 +6\sin^2(x)$

Rearrange the equation to find the value of $\sin^2(x)$ .
$\sin^2(x) = \frac{5}{11}$

Take the square root of both sides of the equation.
$\sin(x) = \pm\sqrt{\frac{5}{11}}$

Calculate the value of $x$ for $-\pi \leq x < \pi$ .
$x = \sin^{-}\left(\pm\sqrt{\frac{5}{11}}\right)$

Calculate the values of $x$ to3 decimal places.
$x \approx \pm0.857, \pm.285$ The solutions are $x \approx0.857, -0.857,.285, -.285$ .