91Ó°ÊÓ

When working with trigonometric equations, the double-angle identity is a powerful tool that simplifies expressions involving angles that are twice another angle. The double-angle identity for cosine is:\[ \ \ \cos 2x = 2 \cos^2 x - 1 \ \ \]In our exercise, we used this identity to rewrite the term \( \cos 2x \), which helped us transform the complex equation into a more manageable form. Knowing and applying these identities can greatly simplify trigonometric equations and make solving them more straightforward.

Replacing \( \cos 2x \) with \( 2 \cos^2 x - 1 \) in the original equation allows us to work with a single type of trigonometric function. This substitution can be crucial for simplifying and solving the equation.

In this specific problem, the transformation leads to:

• Original: \( 5 \cos 2x + \sin x =4 \)
• After substitution: \( 5(2 \cos^2 x - 1) + \sin x = 4 \)

This step is vital as it converts the trigonometric identities into an equation that we can more easily solve.

The quadratic formula is a staple in algebra that allows us to find the roots of any quadratic equation of the form \( ax^2 + bx + c = 0 \). The formula is:\[ \quad\quad x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \ \]In this exercise, after substituting \( 1 - \sin^2 x \) for \( \cos^2 x \), we end up with a quadratic equation in terms of \( \sin x \) (denoted as \( y \) temporarily):\[ -10 \sin^2 x + \sin x + 1 = 0 \ \]
This reduces to:\[ 10 \sin^2 x - \sin x - 1 = 0 \ \]Here, \( a = 10 \), \( b = -1 \), and \( c = -1 \). Plugging these values into the quadratic formula, we solve for \( y \):\[ y = \frac{1 \pm \sqrt{1 + 40}}{20} = \frac{1 \pm \sqrt{41}}{20} \ \]This results in two potential solutions. Evaluating each in the context of the sine function's range (\(-1 \leq \sin x \leq 1\)), we find that only one solution is valid: \( \frac{1 - \sqrt{41}}{20} \).

Thus, using the quadratic formula is essential in our method to solve for \( \sin x \) and verify which solutions are feasible within the problem's constraints.

The sine function is fundamental in trigonometry. It varies between -1 and 1 and is periodic with a period of \( 2 \pi \). Understanding its behavior is crucial for solving trigonometric equations.

In our problem, after manipulating the original trigonometric equation, we ended up with a scenario where we needed to solve for \( \sin x \). Because the sine function adheres to specific constraints \( (-1 \leq \sin x \leq 1) \), not all solutions derived from our quadratic formula are valid.

Our equation yielded two potential solutions for \( y \):

• \( \ y_1 = \frac{1 + \sqrt{41}}{20} \)
• \(y_2 = \frac{1 - \sqrt{41}}{20} \)

Only \(y_2 = \frac{1 - \sqrt{41}}{20} \) is within the range of the sine function. Hence, we only consider that solution:

To find \(x \) from \(\sin x = \frac{1 - \sqrt{41}}{20} \), we utilize the arcsine function. Generally, the solutions for \(\sin x = y \) happen at:

• \( x = \arcsin y \)
• \( x = \pi - \arcsin y \)

Thus, confirming the possible values within our given range \([0, 2\pi)\), it is clear that wrapping trigonometric and algebraic identities and understanding the sine function's properties critically contributes to solving these equations correctly.