QuestionSolve the equations: a) $\frac{2}{\sin x}-\frac{\sin x}{1+\cos x}=\frac{1+\cos x}{\sin x}$ ; c) $\operatorname{tg} x\left(\operatorname{cotg}^{2} x-1\right)=\operatorname{cotg} x\left(1-\operatorname{tg}^{2} x\right)$ .

Studdy Solution

STEP 1

Assumptions1. All the trigonometric functions are defined for the given values of $x$ . . The operations of addition, subtraction, multiplication, and division are all valid for the given values of $x$ .
Let's start with the first equationa) $\frac{}{\sin x}-\frac{\sin x}{1+\cos x}=\frac{1+\cos x}{\sin x}$

STEP 2

To simplify the equation, we will first find a common denominator for the left side of the equation. The common denominator is $\sin x(1+\cos x)$ .
$\frac{2(1+\cos x)}{\sin x(1+\cos x)}-\frac{\sin^2 x}{\sin x(1+\cos x)}$

STEP 3

Subtract the two fractions on the left side of the equation.
$\frac{2(1+\cos x) - \sin^2 x}{\sin x(1+\cos x)}$

STEP 4

implify the numerator using the Pythagorean identity $\sin^2 x + \cos^2 x =1$ .
$\frac{2 - \sin^2 x +2\cos x}{\sin x(1+\cos x)}$

STEP 5

Now, simplify the right side of the equation by multiplying the numerator and the denominator by $(1+\cos x)$ .
$\frac{(1+\cos x)^2}{\sin x(1+\cos x)}$

STEP 6

Now, equate the simplified left side and right side of the equation.
$\frac{2 - \sin^2 x +2\cos x}{\sin x(1+\cos x)} = \frac{(1+\cos x)^2}{\sin x(1+\cos x)}$

STEP 7

Since the denominators on both sides are the same, we can equate the numerators.
$2 - \sin^2 x +2\cos x = (1+\cos x)^2$

STEP 8

implify the right side of the equation.
$2 - \sin^2 x +2\cos x =1 +2\cos x + \cos^2 x$

STEP 9

Rearrange the equation to one side to get a quadratic equation in terms of $\cos x$ .
$\cos^2 x -2\cos x + \sin^2 x - =$

STEP 10

Substitute $\sin^2 x$ with $- \cos^2 x$ .
$2\cos^2 x -2\cos x =0$

STEP 11

Factor out $\cos x$ .
$\cos x(\cos x -) =0$

STEP 12

Set each factor equal to zero and solve for $x$ .
$\cos x =0 \quad \text{or} \quad \cos x =$ These are the solutions for the first equation.
Now, let's move on to the second equationc) $\operatorname{tg} x\left(\operatorname{cotg}^{2} x-\right)=\operatorname{cotg} x\left(-\operatorname{tg}^{2} x\right)$

STEP 13

We can simplify the equation by using the reciprocal identity $\operatorname{cotg} x = \frac{}{\operatorname{tg} x}$ .
$\operatorname{tg} x\left(\frac{}{\operatorname{tg}^2 x}-\right)=\frac{}{\operatorname{tg} x}\left(-\operatorname{tg}^{2} x\right)$

STEP 14

Multiply each term inside the parentheses by $\operatorname{tg}^2 x$ .
$\operatorname{tg} x( - \operatorname{tg}^2 x) = \frac{ - \operatorname{tg}^2 x}{\operatorname{tg} x}$

STEP 15

Multiply both sides by $\operatorname{tg} x$ to get rid of the denominator on the right side.
$\operatorname{tg}^2 x( - \operatorname{tg}^2 x) = - \operatorname{tg}^2 x$

STEP 16

Rearrange the equation to one side to get a quadratic equation in terms of $\operatorname{tg} x$ .
$\operatorname{tg}^4 x - \operatorname{tg}^2 x =0$

STEP 17

Factor out $\operatorname{tg}^2 x$ .
$\operatorname{tg}^2 x(\operatorname{tg}^2 x -) =0$

STEP 18

Set each factor equal to zero and solve for $x$ .
$\operatorname{tg} x =0 \quad \text{or} \quad \operatorname{tg} x = \pm$ These are the solutions for the second equation.

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QuestionSolve the equations: a) $\frac{2}{\sin x}-\frac{\sin x}{1+\cos x}=\frac{1+\cos x}{\sin x}$ ; c) $\operatorname{tg} x\left(\operatorname{cotg}^{2} x-1\right)=\operatorname{cotg} x\left(1-\operatorname{tg}^{2} x\right)$ .

Studdy Solution

STEP 1

STEP 2

To simplify the equation, we will first find a common denominator for the left side of the equation. The common denominator is $\sin x(1+\cos x)$ .
$\frac{2(1+\cos x)}{\sin x(1+\cos x)}-\frac{\sin^2 x}{\sin x(1+\cos x)}$

STEP 3

Subtract the two fractions on the left side of the equation.
$\frac{2(1+\cos x) - \sin^2 x}{\sin x(1+\cos x)}$

STEP 4

implify the numerator using the Pythagorean identity $\sin^2 x + \cos^2 x =1$ .
$\frac{2 - \sin^2 x +2\cos x}{\sin x(1+\cos x)}$

STEP 5

Now, simplify the right side of the equation by multiplying the numerator and the denominator by $(1+\cos x)$ .
$\frac{(1+\cos x)^2}{\sin x(1+\cos x)}$

STEP 6

Now, equate the simplified left side and right side of the equation.
$\frac{2 - \sin^2 x +2\cos x}{\sin x(1+\cos x)} = \frac{(1+\cos x)^2}{\sin x(1+\cos x)}$

STEP 7

Since the denominators on both sides are the same, we can equate the numerators.
$2 - \sin^2 x +2\cos x = (1+\cos x)^2$

STEP 8

implify the right side of the equation.
$2 - \sin^2 x +2\cos x =1 +2\cos x + \cos^2 x$

STEP 9

Rearrange the equation to one side to get a quadratic equation in terms of $\cos x$ .
$\cos^2 x -2\cos x + \sin^2 x - =$

STEP 10

Substitute $\sin^2 x$ with $- \cos^2 x$ .
$2\cos^2 x -2\cos x =0$

STEP 11

Factor out $\cos x$ .
$\cos x(\cos x -) =0$

STEP 12

STEP 13

We can simplify the equation by using the reciprocal identity $\operatorname{cotg} x = \frac{}{\operatorname{tg} x}$ .
$\operatorname{tg} x\left(\frac{}{\operatorname{tg}^2 x}-\right)=\frac{}{\operatorname{tg} x}\left(-\operatorname{tg}^{2} x\right)$

STEP 14

Multiply each term inside the parentheses by $\operatorname{tg}^2 x$ .
$\operatorname{tg} x( - \operatorname{tg}^2 x) = \frac{ - \operatorname{tg}^2 x}{\operatorname{tg} x}$

STEP 15

Multiply both sides by $\operatorname{tg} x$ to get rid of the denominator on the right side.
$\operatorname{tg}^2 x( - \operatorname{tg}^2 x) = - \operatorname{tg}^2 x$

STEP 16

Rearrange the equation to one side to get a quadratic equation in terms of $\operatorname{tg} x$ .
$\operatorname{tg}^4 x - \operatorname{tg}^2 x =0$

STEP 17

Factor out $\operatorname{tg}^2 x$ .
$\operatorname{tg}^2 x(\operatorname{tg}^2 x -) =0$

STEP 18

Set each factor equal to zero and solve for $x$ .
$\operatorname{tg} x =0 \quad \text{or} \quad \operatorname{tg} x = \pm$ These are the solutions for the second equation.

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Studdy Solution

STEP 1

STEP 2

STEP 3

Subtract the two fractions on the left side of the equation.2(1+cos⁡x)−sin⁡2xsin⁡x(1+cos⁡x)\frac{2(1+\cos x) - \sin^2 x}{\sin x(1+\cos x)}sinx(1+cosx)2(1+cosx)−sin2x​

STEP 4

implify the numerator using the Pythagorean identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x =1sin2x+cos2x=1.2−sin⁡2x+2cos⁡xsin⁡x(1+cos⁡x)\frac{2 - \sin^2 x +2\cos x}{\sin x(1+\cos x)}sinx(1+cosx)2−sin2x+2cosx​

STEP 5

Now, simplify the right side of the equation by multiplying the numerator and the denominator by (1+cos⁡x)(1+\cos x)(1+cosx).(1+cos⁡x)2sin⁡x(1+cos⁡x)\frac{(1+\cos x)^2}{\sin x(1+\cos x)}sinx(1+cosx)(1+cosx)2​

STEP 6

Now, equate the simplified left side and right side of the equation.2−sin⁡2x+2cos⁡xsin⁡x(1+cos⁡x)=(1+cos⁡x)2sin⁡x(1+cos⁡x)\frac{2 - \sin^2 x +2\cos x}{\sin x(1+\cos x)} = \frac{(1+\cos x)^2}{\sin x(1+\cos x)}sinx(1+cosx)2−sin2x+2cosx​=sinx(1+cosx)(1+cosx)2​

STEP 7

Since the denominators on both sides are the same, we can equate the numerators.2−sin⁡2x+2cos⁡x=(1+cos⁡x)22 - \sin^2 x +2\cos x = (1+\cos x)^22−sin2x+2cosx=(1+cosx)2

STEP 8

implify the right side of the equation.2−sin⁡2x+2cos⁡x=1+2cos⁡x+cos⁡2x2 - \sin^2 x +2\cos x =1 +2\cos x + \cos^2 x2−sin2x+2cosx=1+2cosx+cos2x

STEP 9

Rearrange the equation to one side to get a quadratic equation in terms of cos⁡x\cos xcosx.cos⁡2x−2cos⁡x+sin⁡2x−=\cos^2 x -2\cos x + \sin^2 x - =cos2x−2cosx+sin2x−=

STEP 10

Substitute sin⁡2x\sin^2 xsin2x with −cos⁡2x - \cos^2 x−cos2x.2cos⁡2x−2cos⁡x=02\cos^2 x -2\cos x =02cos2x−2cosx=0

STEP 11

Factor out cos⁡x\cos xcosx.cos⁡x(cos⁡x−)=0\cos x(\cos x -) =0cosx(cosx−)=0

STEP 12

STEP 13

STEP 14

Multiply each term inside the parentheses by tg⁡2x\operatorname{tg}^2 xtg2x.tg⁡x(−tg⁡2x)=−tg⁡2xtg⁡x\operatorname{tg} x( - \operatorname{tg}^2 x) = \frac{ - \operatorname{tg}^2 x}{\operatorname{tg} x}tgx(−tg2x)=tgx−tg2x​

STEP 15

Multiply both sides by tg⁡x\operatorname{tg} xtgx to get rid of the denominator on the right side.tg⁡2x(−tg⁡2x)=−tg⁡2x\operatorname{tg}^2 x( - \operatorname{tg}^2 x) = - \operatorname{tg}^2 xtg2x(−tg2x)=−tg2x

STEP 16

Rearrange the equation to one side to get a quadratic equation in terms of tg⁡x\operatorname{tg} xtgx.tg⁡4x−tg⁡2x=0\operatorname{tg}^4 x - \operatorname{tg}^2 x =0tg4x−tg2x=0

STEP 17

Factor out tg⁡2x\operatorname{tg}^2 xtg2x.tg⁡2x(tg⁡2x−)=0\operatorname{tg}^2 x(\operatorname{tg}^2 x -) =0tg2x(tg2x−)=0

STEP 18

Set each factor equal to zero and solve for xxx.tg⁡x=0ortg⁡x=±\operatorname{tg} x =0 \quad \text{or} \quad \operatorname{tg} x = \pmtgx=0ortgx=±These are the solutions for the second equation.

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Subtract the two fractions on the left side of the equation.
$\frac{2(1+\cos x) - \sin^2 x}{\sin x(1+\cos x)}$

implify the numerator using the Pythagorean identity $\sin^2 x + \cos^2 x =1$ .
$\frac{2 - \sin^2 x +2\cos x}{\sin x(1+\cos x)}$

Now, simplify the right side of the equation by multiplying the numerator and the denominator by $(1+\cos x)$ .
$\frac{(1+\cos x)^2}{\sin x(1+\cos x)}$

Now, equate the simplified left side and right side of the equation.
$\frac{2 - \sin^2 x +2\cos x}{\sin x(1+\cos x)} = \frac{(1+\cos x)^2}{\sin x(1+\cos x)}$

Since the denominators on both sides are the same, we can equate the numerators.
$2 - \sin^2 x +2\cos x = (1+\cos x)^2$

implify the right side of the equation.
$2 - \sin^2 x +2\cos x =1 +2\cos x + \cos^2 x$

Rearrange the equation to one side to get a quadratic equation in terms of $\cos x$ .
$\cos^2 x -2\cos x + \sin^2 x - =$

Substitute $\sin^2 x$ with $- \cos^2 x$ .
$2\cos^2 x -2\cos x =0$

Factor out $\cos x$ .
$\cos x(\cos x -) =0$

Multiply each term inside the parentheses by $\operatorname{tg}^2 x$ .
$\operatorname{tg} x( - \operatorname{tg}^2 x) = \frac{ - \operatorname{tg}^2 x}{\operatorname{tg} x}$

Multiply both sides by $\operatorname{tg} x$ to get rid of the denominator on the right side.
$\operatorname{tg}^2 x( - \operatorname{tg}^2 x) = - \operatorname{tg}^2 x$

Rearrange the equation to one side to get a quadratic equation in terms of $\operatorname{tg} x$ .
$\operatorname{tg}^4 x - \operatorname{tg}^2 x =0$

Factor out $\operatorname{tg}^2 x$ .
$\operatorname{tg}^2 x(\operatorname{tg}^2 x -) =0$

Set each factor equal to zero and solve for $x$ .
$\operatorname{tg} x =0 \quad \text{or} \quad \operatorname{tg} x = \pm$ These are the solutions for the second equation.