Solve The Equation:$\[ \frac{1}{\sin X + 1} - \frac{2}{\cos X} = \frac{\sin X + 1}{\sin X - 1} \\]

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Introduction

Trigonometric equations are a fundamental aspect of mathematics, and solving them requires a deep understanding of trigonometric functions and identities. In this article, we will focus on solving a specific trigonometric equation that involves sine and cosine functions. The equation is given as:

1sin⁑x+1βˆ’2cos⁑x=sin⁑x+1sin⁑xβˆ’1{ \frac{1}{\sin x + 1} - \frac{2}{\cos x} = \frac{\sin x + 1}{\sin x - 1} }

Understanding the Equation

The given equation involves three trigonometric functions: sine, cosine, and the reciprocal of sine. To solve this equation, we need to simplify it and isolate the variable x. The equation can be rewritten as:

1sin⁑x+1βˆ’2cos⁑xβˆ’sin⁑x+1sin⁑xβˆ’1=0{ \frac{1}{\sin x + 1} - \frac{2}{\cos x} - \frac{\sin x + 1}{\sin x - 1} = 0 }

Simplifying the Equation

To simplify the equation, we can start by finding a common denominator for the three fractions. The common denominator is (sin x + 1)(cos x)(sin x - 1). Multiplying each fraction by the necessary factors to achieve this common denominator, we get:

(cosx)(sinxβˆ’1)βˆ’2(sinx+1)(sinxβˆ’1)βˆ’(sinx+1)(cosx)(sinx+1)(cosx)(sinxβˆ’1)=0{ \frac{(cos x)(sin x - 1) - 2(sin x + 1)(sin x - 1) - (sin x + 1)(cos x)}{(sin x + 1)(cos x)(sin x - 1)} = 0 }

Expanding and Simplifying

Expanding the numerator of the fraction, we get:

cosxsinxβˆ’cosxβˆ’2sin2xβˆ’2(sinx+1)(cosx)(sinxβˆ’1)=0{ \frac{cos x sin x - cos x - 2 sin^2 x - 2}{(sin x + 1)(cos x)(sin x - 1)} = 0 }

Using Trigonometric Identities

To simplify the equation further, we can use the trigonometric identity sin^2 x + cos^2 x = 1. Rearranging this identity, we get:

cos2x=1βˆ’sin2x{ cos^2 x = 1 - sin^2 x }

Substituting this expression into the equation, we get:

cosxsinxβˆ’cosxβˆ’2sin2xβˆ’2(sinx+1)(cosx)(sinxβˆ’1)=0{ \frac{cos x sin x - cos x - 2 sin^2 x - 2}{(sin x + 1)(cos x)(sin x - 1)} = 0 }

Solving for x

To solve for x, we need to isolate the variable x. We can start by factoring the numerator of the fraction:

(cosxβˆ’2sinxβˆ’2)(sinxβˆ’1)(sinx+1)(cosx)(sinxβˆ’1)=0{ \frac{(cos x - 2 sin x - 2)(sin x - 1)}{(sin x + 1)(cos x)(sin x - 1)} = 0 }

Canceling Common Factors

Canceling the common factor (sin x - 1) from the numerator and denominator, we get:

cosxβˆ’2sinxβˆ’2(sinx+1)(cosx)=0{ \frac{cos x - 2 sin x - 2}{(sin x + 1)(cos x)} = 0 }

Solving for x

To solve for x, we need to set the numerator of the fraction equal to zero:

cosxβˆ’2sinxβˆ’2=0{ cos x - 2 sin x - 2 = 0 }

Using Trigonometric Identities

To solve for x, we can use the trigonometric identity cos x = 1 - 2 sin^2 x. Substituting this expression into the equation, we get:

1βˆ’2sin2xβˆ’2sinxβˆ’2=0{ 1 - 2 sin^2 x - 2 sin x - 2 = 0 }

Rearranging the Equation

Rearranging the equation, we get:

2sin2x+2sinx+1=0{ 2 sin^2 x + 2 sin x + 1 = 0 }

Solving the Quadratic Equation

The equation is a quadratic equation in terms of sin x. We can solve it using the quadratic formula:

sinx=βˆ’bΒ±b2βˆ’4ac2a{ sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} }

Substituting Values

Substituting the values a = 2, b = 2, and c = 1 into the quadratic formula, we get:

sinx=βˆ’2Β±22βˆ’4(2)(1)2(2){ sin x = \frac{-2 \pm \sqrt{2^2 - 4(2)(1)}}{2(2)} }

Simplifying the Expression

Simplifying the expression, we get:

sinx=βˆ’2Β±βˆ’44{ sin x = \frac{-2 \pm \sqrt{-4}}{4} }

Using Complex Numbers

Since the expression under the square root is negative, we can use complex numbers to simplify it. Let's denote the complex number as i = √(-1). Then, we can rewrite the expression as:

sinx=βˆ’2Β±2i4{ sin x = \frac{-2 \pm 2i}{4} }

Simplifying the Expression

Simplifying the expression, we get:

sinx=βˆ’1Β±i2{ sin x = \frac{-1 \pm i}{2} }

Finding the Values of x

To find the values of x, we need to take the inverse sine of both sides of the equation:

x=sinβ‘βˆ’1(βˆ’1Β±i2){ x = \sin^{-1} \left( \frac{-1 \pm i}{2} \right) }

Using the Inverse Sine Function

The inverse sine function is a multivalued function, which means that it has multiple values for a given input. To find the values of x, we need to use the principal value of the inverse sine function, which is defined as:

sinβ‘βˆ’1x=Ο€2βˆ’tanβ‘βˆ’1(x1){ \sin^{-1} x = \frac{\pi}{2} - \tan^{-1} \left( \frac{x}{1} \right) }

Substituting Values

Substituting the values x = (-1 Β± i)/2 into the inverse sine function, we get:

x=Ο€2βˆ’tanβ‘βˆ’1(βˆ’1Β±i2){ x = \frac{\pi}{2} - \tan^{-1} \left( \frac{-1 \pm i}{2} \right) }

Simplifying the Expression

Simplifying the expression, we get:

x=Ο€2βˆ’tanβ‘βˆ’1(βˆ’1Β±i2){ x = \frac{\pi}{2} - \tan^{-1} \left( \frac{-1 \pm i}{2} \right) }

Finding the Values of x

To find the values of x, we need to evaluate the inverse tangent function:

tanβ‘βˆ’1(βˆ’1Β±i2){ \tan^{-1} \left( \frac{-1 \pm i}{2} \right) }

Using the Inverse Tangent Function

The inverse tangent function is a multivalued function, which means that it has multiple values for a given input. To find the values of x, we need to use the principal value of the inverse tangent function, which is defined as:

tanβ‘βˆ’1x=12ln⁑(1+x1βˆ’x){ \tan^{-1} x = \frac{1}{2} \ln \left( \frac{1 + x}{1 - x} \right) }

Substituting Values

Substituting the values x = (-1 Β± i)/2 into the inverse tangent function, we get:

tanβ‘βˆ’1(βˆ’1Β±i2)=12ln⁑(1+βˆ’1Β±i21βˆ’βˆ’1Β±i2){ \tan^{-1} \left( \frac{-1 \pm i}{2} \right) = \frac{1}{2} \ln \left( \frac{1 + \frac{-1 \pm i}{2}}{1 - \frac{-1 \pm i}{2}} \right) }

Simplifying the Expression

Simplifying the expression, we get:

tanβ‘βˆ’1(βˆ’1Β±i2)=12ln⁑(1βˆ“i23βˆ“i2){ \tan^{-1} \left( \frac{-1 \pm i}{2} \right) = \frac{1}{2} \ln \left( \frac{\frac{1 \mp i}{2}}{\frac{3 \mp i}{2}} \right) }

Simplifying the Expression

Simplifying the expression, we get:

tanβ‘βˆ’1(βˆ’1Β±i2)=12ln⁑(1βˆ“i3βˆ“i){ \tan^{-1} \left( \frac{-1 \pm i}{2} \right) = \frac{1}{2} \ln \left( \frac{1 \mp i}{3 \mp i} \right) }

Simplifying the Expression

Simplifying the expression, we get:

tanβ‘βˆ’1(βˆ’1Β±i2)=12ln⁑(1βˆ“i3βˆ“i){ \tan^{-1} \left( \frac{-1 \pm i}{2} \right) = \frac{1}{2} \ln \left( \frac{1 \mp i}{3 \mp i} \right) }

Finding the Values of x

To find the values of x, we need to evaluate the logarithmic expression:

12ln⁑(1βˆ“i3βˆ“i){ \frac{1}{2} \ln \left( \frac{1 \mp i}{3 \mp i} \right) }

Using the Properties of Logarithms

The logarithmic expression can be simplified using the properties of logarithms. Specifically, we can use the property:

ln⁑(ab)=ln⁑aβˆ’ln⁑b{ \ln \left( \frac{a}{b} \right) = \ln a - \ln b }

Substituting Values

Substituting the values a = (1 Β± i) and b = (3 Β± i) into the logarithmic expression, we get:

12ln⁑(1βˆ“i3βˆ“i)=12(ln⁑(1Β±i)βˆ’ln⁑(3Β±i)){ \frac{1}{2} \ln \left( \frac{1 \mp i}{3 \mp i} \right) = \frac{1}{2} (\ln (1 \pm i) - \ln (3 \pm i)) }

Simplifying the Expression

Simplifying

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Q: What is the main goal of solving the trigonometric equation?

A: The main goal of solving the trigonometric equation is to find the values of x that satisfy the equation.

Q: What are the steps involved in solving the trigonometric equation?

A: The steps involved in solving the trigonometric equation are:

  1. Simplifying the equation
  2. Using trigonometric identities
  3. Solving for x
  4. Using complex numbers
  5. Finding the values of x

Q: What is the significance of using complex numbers in solving the trigonometric equation?

A: Complex numbers are used in solving the trigonometric equation because the expression under the square root is negative. This requires the use of complex numbers to simplify the expression.

Q: How do you find the values of x using the inverse sine function?

A: To find the values of x using the inverse sine function, you need to take the inverse sine of both sides of the equation:

x=sinβ‘βˆ’1(βˆ’1Β±i2){ x = \sin^{-1} \left( \frac{-1 \pm i}{2} \right) }

Q: What is the principal value of the inverse sine function?

A: The principal value of the inverse sine function is defined as:

sinβ‘βˆ’1x=Ο€2βˆ’tanβ‘βˆ’1(x1){ \sin^{-1} x = \frac{\pi}{2} - \tan^{-1} \left( \frac{x}{1} \right) }

Q: How do you simplify the expression using the properties of logarithms?

A: To simplify the expression using the properties of logarithms, you can use the property:

ln⁑(ab)=ln⁑aβˆ’ln⁑b{ \ln \left( \frac{a}{b} \right) = \ln a - \ln b }

Q: What is the final answer to the trigonometric equation?

A: The final answer to the trigonometric equation is:

x=Ο€2βˆ’tanβ‘βˆ’1(βˆ’1Β±i2){ x = \frac{\pi}{2} - \tan^{-1} \left( \frac{-1 \pm i}{2} \right) }

Q: What are the possible values of x?

A: The possible values of x are:

x=Ο€2βˆ’tanβ‘βˆ’1(βˆ’1+i2){ x = \frac{\pi}{2} - \tan^{-1} \left( \frac{-1 + i}{2} \right) }

and

x=Ο€2βˆ’tanβ‘βˆ’1(βˆ’1βˆ’i2){ x = \frac{\pi}{2} - \tan^{-1} \left( \frac{-1 - i}{2} \right) }

Q: How do you evaluate the inverse tangent function?

A: To evaluate the inverse tangent function, you can use the property:

tanβ‘βˆ’1x=12ln⁑(1+x1βˆ’x){ \tan^{-1} x = \frac{1}{2} \ln \left( \frac{1 + x}{1 - x} \right) }

Q: What is the significance of the inverse tangent function in solving the trigonometric equation?

A: The inverse tangent function is used in solving the trigonometric equation to find the values of x. It is a multivalued function, which means that it has multiple values for a given input.

Q: How do you find the values of x using the inverse tangent function?

A: To find the values of x using the inverse tangent function, you need to evaluate the expression:

tanβ‘βˆ’1(βˆ’1Β±i2){ \tan^{-1} \left( \frac{-1 \pm i}{2} \right) }

Q: What are the possible values of x using the inverse tangent function?

A: The possible values of x using the inverse tangent function are:

x=Ο€2βˆ’tanβ‘βˆ’1(βˆ’1+i2){ x = \frac{\pi}{2} - \tan^{-1} \left( \frac{-1 + i}{2} \right) }

and

x=Ο€2βˆ’tanβ‘βˆ’1(βˆ’1βˆ’i2){ x = \frac{\pi}{2} - \tan^{-1} \left( \frac{-1 - i}{2} \right) }

Q: What is the final answer to the trigonometric equation using the inverse tangent function?

A: The final answer to the trigonometric equation using the inverse tangent function is:

x=Ο€2βˆ’tanβ‘βˆ’1(βˆ’1Β±i2){ x = \frac{\pi}{2} - \tan^{-1} \left( \frac{-1 \pm i}{2} \right) }