Question

Rewrite each equation in terms of sine only. Then, solve algebraically for \(0 \leq x<2 \pi\). a) \(\cos 2 x-3 \sin x=2\) b) \(2 \cos ^{2} x-3 \sin x-3=0\) c) \(3 \csc x-\sin x=2\) d) \(\tan ^{2} x+2=0\)

Short answer

a) \( x = \frac{\pi}{2}, \frac{\pi}{6}, \frac{5 \pi}{6} \) b) Same solution as a) c) \( x = \frac{\pi}{2} \) d) No solution.

Step-by-step solution

Rewrite \(\cos 2x\) using sine

Use the double-angle formula: \[ \cos 2x = 1 - 2 \sin^2 x \] for part a).

Substitute and simplify for part a)

Substitute the double-angle identity into the equation: \[ 1 - 2 \sin^2 x - 3 \sin x = 2 \] Simplify: \[ -2 \sin^2 x - 3 \sin x - 1 = 0 \] Let \(u = \sin x\): \[ -2u^2 - 3u - 1 = 0 \]

Solve the quadratic equation for \(u\)

Solve for \(u = \sin x \): \[ u = \frac{-3 \pm \sqrt{9 - 8}}{2 (-2)} = \frac{-3 \pm 1}{-4} = 1, \frac{1}{2} \] Therefore: \[ \sin x = 1 \Rightarrow x = \frac{\pi}{2} \] \[ \sin x = \frac{1}{2} \Rightarrow x = \frac{\pi}{6}, \frac{5 \pi}{6} \]

Rewrite \(\cos^2 x\) using sine

Use the Pythagorean identity: \[ \cos^2 x = 1 - \sin^2 x \] for part b).

Substitute and simplify for part b)

Substitute the Pythagorean identity into the equation: \[ 2(1-\sin^2 x) - 3 \sin x - 3 = 0 \] Simplify: \[ 2 - 2 \sin^2 x - 3 \sin x - 3 = 0 \] Let \(u = \sin x\): \[ -2u^2 - 3u - 1 = 0 \] Solve: Same solution as part a).

Rewrite \(\csc x\) using sine

Use the identity: \[ \csc x = \frac{1}{\sin x} \] for part c).

Substitute and simplify for part c)

Substitute the identity into the equation: \[ 3 \frac{1}{\sin x} - \sin x = 2 \] Multiply through by \(\sin x\): \[ 3 - \sin^2 x = 2 \sin x \] Let \(u = \sin x\): \[ 3 - u^2 - 2u = 0 \]

Solve the quadratic equation for \(u\)

Rearrange and solve: \[ u^2 + 2u - 3 = 0 \] Factoring: \[ (u + 3)(u - 1) = 0 \Rightarrow u = -3, 1 \] Since \(u = \sin x\), valid solutions are: \[ \sin x = 1 \Rightarrow x = \frac{\pi}{2} \]

Solve \(\tan^2 x + 2 = 0\) for part d)

Recognize that the equation has no solution since \(\tan^2 x \) is always non-negative and cannot satisfy an equation where its value plus 2 equals zero.

Key concepts

Trigonometric Identities

Trigonometric identities are true for all values of the variable(s) involved where both sides of the equation are defined. They help simplify complex trigonometric expressions and solve trigonometric equations. Some common trigonometric identities include:

• Pythagorean Identities: \(\text{sin}^2 x + \text{cos}^2 x = 1\)
• Double-Angle Formulas: \(\text{cos}(2x) = 1 - 2\text{sin}^2 x\)
• Reciprocal Identities: \(\text{csc} x = \frac{1}{\text{sin} x}\)

In the exercise, we used these identities to rewrite and simplify trigonometric equations. Understanding and being able to apply these identities is crucial for solving more complex problems.

Solving Quadratic Equations

Solving quadratic equations involves finding the values of the variable that satisfy the equation. A quadratic equation is of the form \(ax^2 + bx + c = 0\). To solve it, you can use methods like:

• Factoring: Rewriting the equation as a product of simpler binomials.
• Quadratic Formula: \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
• Completing the Square: Manipulating the equation so it expresses a perfect square trinomial.

In the exercise, we transformed the trigonometric equations into quadratic equations by using identities. For example, substituting \(u = \text{sin} x\) and solving \(-2u^2 - 3u - 1 = 0\) helped find the solutions for \text{sin} x\.

Double-Angle Formulas

Double-angle formulas express trigonometric functions of double angles (like \(2x\)) in terms of single angles (like \(x\)). They are particularly useful in simplifying expressions and solving equations. The two most commonly used double-angle formulas are:

• \(\text{cos}(2x) = \cos^2 x - \sin^2 x = 2\text{cos}^2 x - 1 = 1 - 2\text{sin}^2 x\)
• \(\text{sin}(2x) = 2\text{sin} x \text{cos} x\)

In part a) of the exercise, we applied the double-angle formula for cosine: \( \text{cos}(2x) = 1 - 2 \text{sin}^2 x\). This helped rewrite the equation in terms of \text{sin} x\ so we could proceed with solving it algebraically.

Pythagorean Identities

Pythagorean identities relate the squares of trigonometric functions. They derive from the Pythagorean theorem in a unit circle where the radius is 1. The primary Pythagorean identity is:

• \(\text{sin}^2 x + \text{cos}^2 x = 1\)

Other derived identities include:

• \(1 + \text{tan}^2 x = \sec^2 x\)
• \(1 + \text{cot}^2 x = \csc^2 x\)

In part b) of the exercise, we used the Pythagorean identity to rewrite \( \text{cos}^2 x\) as \(1 - \text{sin}^2 x\). This simplification allowed us to transform the given equation into a solvable quadratic equation in terms of \text{sin} x\.