Solving the Equation sin x - cos x - 1/√2 0 within the Interval [0, 2π]

Trigonometric equations are fundamental in mathematics, particularly in calculus, physics, and engineering. One such equation is sin x - cos x - 1/√2 0. This article will guide you through solving this equation for x in the interval [0, 2π]. We will present step-by-step solutions, highlighting each key step and providing proofs for the derived solutions.

Introduction to the Equation

The given equation is sin x - cos x - 1/√2 0. Let's first rewrite this equation for clarity:

sin x - cos x 1/√2

To solve this equation, we will employ algebraic manipulation, specifically squaring both sides, to simplify the expression and find possible solutions for x within the specified interval [0, 2π].

Squaring Both Sides

Our first step involves squaring both sides of the equation:

sin x - cos x2 (1/√2)2

Expanding the left side using the identity sin^2 x - cos^2 x - 2 sin x cos x, we get:

sin^2 x - cos^2 x - 2 sin x cos x 1/2

Next, we recognize the Pythagorean identity 1 - cos 2x and substitute it into our equation:

1 - cos 2x - 2 sin x cos x 1/2

Recalling that 2 sin x cos x sin 2x, we can rewrite the equation as:

1 - cos 2x - sin 2x 1/2

Rearranging the Equation

Subtract 1/2 from both sides to isolate the trigonometric terms:

1/2 - cos 2x - sin 2x 0

Let's isolate cos 2x and sin 2x:

cos 2x sin 2x 1/2

To proceed further, we square both sides again to convert it into a more manageable form:

(cos 2x sin 2x)2 (1/2)2

Expanding the left side, we get:

cos^2 2x 2 sin 2x cos 2x sin^2 2x 1/4

Using the Pythagorean identity sin^2 2x cos^2 2x 1, we simplify the left side:

1 2 sin 2x cos 2x 1/4

Recalling that 2 sin 2x cos 2x sin 4x, we rewrite the equation as:

1 sin 4x 1/4

Subtract 1 from both sides to isolate sin 4x:

sin 4x 1/4 - 1

Simplifying the right side, we get:

sin 4x -3/4

Applying the Inverse Sine Function

To solve for x, we need to use the inverse sine function. Recall that:

4x arcsin(-3/4) nπ

Let's denote arcsin(-3/4) as C where:

C arcsin(-3/4)

Therefore, we have:

x arcsin(-3/4)/4 nπ/4

Now, let's determine the specific solutions in the interval [0, 2π]. We need to find integer values of n such that x falls within the given interval. Let's solve for n:

For n 0:

x arcsin(-3/4)/4

For n 1:

x (arcsin(-3/4) π)/4 ≈ 1.3158 (within range)

For n 2:

x (arcsin(-3/4) 2π)/4 ≈ 3.566 (within range)

For n 3:

x (arcsin(-3/4) 3π)/4 ≈ 5.816 (out of range)

Thus, the solutions within the interval [0, 2π] are:

x (arcsin(-3/4) π)/4, (arcsin(-3/4) 2π)/4

Conclusion

In conclusion, the equation sin x - cos x - 1/√2 0 has solutions within the interval [0, 2π] given by: