Solving sin2x - 2cos^2x 0 for x in the Range from 0° to 180°

Introduction

Whether you're a student of trigonometry or someone working on complex mathematical problems, understanding how to solve equations such as sin2x - 2cos^2x 0 is crucial. In this detailed guide, we will break down the process of solving this equation for x in the range from 0° to 180° using a double angle formula, ensuring you gain a deep understanding of the trigonometric identities and their applications.

Step-by-Step Solution

Let's start with the given equation:

sin 2x - 2{cos^2}x - 1 - 2{cos^2}x  0

Simplifying the equation, we can group the terms involving cosine squared:

sin 2x - 2{cos^2}x - 1 - 2{cos^2}x  0sin 2x - 4{cos^2}x - 1  0

Using the double angle identity for sine, sin 2x 2sinx cosx, we can rewrite the equation:

2sinx cosx - 4{cos^2}x - 1  0

To proceed, let's solve for x in the range from 0° to 180° by isolating the trigonometric functions. We will use the identity that relates sine and cosine:

sin 2x  2sinx cosx2sinx cosx - 4{cos^2}x - 1  0

Notice that the equation can be simplified further by considering cos^2 x, which we can represent as 1 - sin^2 x to form a quadratic equation in terms of sin x.

2sinx cosx - 4{1 - sin^2 x} - 1  02sinx cosx - 4   4sin^2 x - 1  02sinx cosx   4sin^2 x - 5  0

Now, let's simplify further by substituting cosx ±√(1 - sin^2 x):

2sinx(±√(1 - sin^2 x))   4sin^2 x - 5  06sin^4 x - 5sin^2 x - 1  0

Letting u sin^2 x, we have:

6u^2 - 5u - 1  0

Solving this quadratic equation for u, we use the quadratic formula:

u  (-b ± √(b^2 - 4ac)) / 2au  (5 ± √(25   24)) / 12u  (5 ± 7) / 12u  1/2 or -1/6

Since sin^2 x must be non-negative, we take the positive value:

sin^2 x  1/2sin x  ±√(1/2)sin x  ±√(2)/2

Thus, the possible values for x are:

sin x √(2)/2, giving x 45° sin x -√(2)/2, giving x 135°

Therefore, the solutions for x in the given range from 0° to 180° are:

x  45°, 135°

Conclusion

In this detailed solution, we've explored how to solve the equation sin2x - 2cos^2x 0 using the double angle formula and trigonometric identities. The key step involved simplifying the equation to a quadratic in terms of sine squared and solving it using the quadratic formula. The final solutions were found to be x 45° and x 135°. With this understanding, you should be well-equipped to tackle similar trigonometric problems.

References

[1] "Trigonometric Identities" - MathIsFun

[2] "Double Angle Formulae" - MathIsFun

[3] "Solving Quadratic Equations" - Khan Academy