Math Problem Statement

Solve the equation 2 cos² θ° = 1 - sin θ° for 0 ≤ θ ≤ 360°.

Solution

To solve the equation 2cos2θ=1sinθ2 \cos^2 \theta^\circ = 1 - \sin \theta^\circ for 0θ3600 \leq \theta \leq 360^\circ, we can follow these steps:

Step-by-Step Solution

  1. Rewrite cos2θ\cos^2 \theta in terms of sinθ\sin \theta:

    Using the Pythagorean identity cos2θ=1sin2θ\cos^2 \theta = 1 - \sin^2 \theta, substitute this into the equation: 2(1sin2θ)=1sinθ.2(1 - \sin^2 \theta) = 1 - \sin \theta.

  2. Expand and simplify: 22sin2θ=1sinθ.2 - 2 \sin^2 \theta = 1 - \sin \theta. Move all terms to one side to set the equation to zero: 2sin2θsinθ1=0.2 \sin^2 \theta - \sin \theta - 1 = 0.

  3. Solve the quadratic equation in sinθ\sin \theta:

    This is a quadratic equation in terms of sinθ\sin \theta, where a=2a = 2, b=1b = -1, and c=1c = -1: 2sin2θsinθ1=0.2 \sin^2 \theta - \sin \theta - 1 = 0. Use the quadratic formula: sinθ=(1)±(1)242(1)22.\sin \theta = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2}. Simplify inside the square root: sinθ=1±1+84=1±34.\sin \theta = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4}. This gives two solutions: sinθ=1+34=1andsinθ=134=12.\sin \theta = \frac{1 + 3}{4} = 1 \quad \text{and} \quad \sin \theta = \frac{1 - 3}{4} = -\frac{1}{2}.

  4. Find θ\theta for each case:

    • Case 1: sinθ=1\sin \theta = 1

      • θ=90\theta = 90^\circ.
    • Case 2: sinθ=12\sin \theta = -\frac{1}{2}

      • For sinθ=12\sin \theta = -\frac{1}{2}, the angles are θ=210\theta = 210^\circ and θ=330\theta = 330^\circ in the range 0θ3600 \leq \theta \leq 360^\circ.

Final Solutions

The values of θ\theta that satisfy the equation are: θ=90,θ=210,θ=330.\theta = 90^\circ, \quad \theta = 210^\circ, \quad \theta = 330^\circ.

Would you like further details on any step?

Related Questions

  1. How do we apply trigonometric identities to simplify equations?
  2. What other trigonometric equations can be solved by converting to a quadratic form?
  3. How can we verify if a solution to a trigonometric equation is correct?
  4. How do we handle trigonometric equations with restricted domains?
  5. Can this approach be used for equations with functions like tanθ\tan \theta or secθ\sec \theta?

Tip

When solving trigonometric equations, always remember to check if the solutions fall within the given range or interval for the variable.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Trigonometric Identities
Quadratic Equations

Formulas

cos² θ = 1 - sin² θ
Quadratic formula: x = (-b ± √(b² - 4ac)) / 2a

Theorems

Pythagorean identity

Suitable Grade Level

Grades 10-12