Math Problem Statement

Solve the given equation. (a) sin θ = √2 2 (b) √2 cos θ + 1 = 0 (rationalize to get a value we know) (c) tan θ + 1 = 0 (d) sin2 x − sin x − 2 = 0

Solution

Let us solve each equation step by step:


(a) sinθ=22\sin \theta = \frac{\sqrt{2}}{2}

The value sinθ=22\sin \theta = \frac{\sqrt{2}}{2} corresponds to the angles where sine equals this value:

  1. θ=π4\theta = \frac{\pi}{4} or θ=45\theta = 45^\circ in the first quadrant.
  2. θ=3π4\theta = \frac{3\pi}{4} or θ=135\theta = 135^\circ in the second quadrant.

Thus, the solutions are: θ=π4,3π4+2nπ(nZ)\theta = \frac{\pi}{4}, \frac{3\pi}{4} + 2n\pi \quad (n \in \mathbb{Z})


(b) 2cosθ+1=0\sqrt{2} \cos \theta + 1 = 0

Step 1: Isolate cosθ\cos \theta:

2cosθ=1\sqrt{2} \cos \theta = -1 cosθ=12\cos \theta = -\frac{1}{\sqrt{2}}

Step 2: Rationalize:

cosθ=22\cos \theta = -\frac{\sqrt{2}}{2}

Step 3: Solve for θ\theta:

The value cosθ=22\cos \theta = -\frac{\sqrt{2}}{2} corresponds to the angles where cosine is negative:

  1. θ=ππ4=3π4\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4} in the second quadrant.
  2. θ=π+π4=5π4\theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4} in the third quadrant.

Thus, the solutions are: θ=3π4,5π4+2nπ(nZ)\theta = \frac{3\pi}{4}, \frac{5\pi}{4} + 2n\pi \quad (n \in \mathbb{Z})


(c) tanθ+1=0\tan \theta + 1 = 0

Step 1: Isolate tanθ\tan \theta:

tanθ=1\tan \theta = -1

Step 2: Solve for θ\theta:

The value tanθ=1\tan \theta = -1 corresponds to the angles where tangent is negative:

  1. θ=ππ4=3π4\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4} in the second quadrant.
  2. θ=2ππ4=7π4\theta = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4} in the fourth quadrant.

Thus, the solutions are: θ=3π4,7π4+nπ(nZ)\theta = \frac{3\pi}{4}, \frac{7\pi}{4} + n\pi \quad (n \in \mathbb{Z})


(d) sin2xsinx2=0\sin^2 x - \sin x - 2 = 0

This is a quadratic equation in sinx\sin x. Let u=sinxu = \sin x, so the equation becomes:

u2u2=0u^2 - u - 2 = 0

Step 1: Factorize the quadratic equation:

u2u2=(u2)(u+1)=0u^2 - u - 2 = (u - 2)(u + 1) = 0

Step 2: Solve for uu:

u - 2 = 0 \quad \Rightarrow \quad u = 2 \quad \text{(Not possible as \sin x \leq 1)} u+1=0u=1u + 1 = 0 \quad \Rightarrow \quad u = -1

Thus, sinx=1\sin x = -1.

Step 3: Solve for xx:

The value sinx=1\sin x = -1 occurs at: x=3π2+2nπ(nZ)x = \frac{3\pi}{2} + 2n\pi \quad (n \in \mathbb{Z})


Final Solutions:

  • (a) θ=π4,3π4+2nπ\theta = \frac{\pi}{4}, \frac{3\pi}{4} + 2n\pi
  • (b) θ=3π4,5π4+2nπ\theta = \frac{3\pi}{4}, \frac{5\pi}{4} + 2n\pi
  • (c) θ=3π4,7π4+nπ\theta = \frac{3\pi}{4}, \frac{7\pi}{4} + n\pi
  • (d) x=3π2+2nπx = \frac{3\pi}{2} + 2n\pi

Would you like detailed steps for any specific part, or do you have any questions?

Here are 5 follow-up questions to explore:

  1. What are the steps to derive trigonometric values from the unit circle?
  2. How does the periodicity of trigonometric functions affect solutions?
  3. What is the significance of rationalizing terms in trigonometric equations?
  4. How can you graphically verify these solutions?
  5. How does factoring help solve trigonometric quadratic equations?

Tip: Always verify solutions in the interval you are working within, especially for periodic functions like sine and cosine!

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Trigonometric Equations
Quadratic Equations

Formulas

sin(θ) = √2/2 → θ = π/4, 3π/4
cos(θ) = -1/√2 → θ = 3π/4, 5π/4
tan(θ) = -1 → θ = 3π/4, 7π/4
Factoring sin²(x) − sin(x) − 2 = 0

Theorems

Unit Circle Theorem
Periodic Properties of Trigonometric Functions
Quadratic Equation Theorem

Suitable Grade Level

Grades 10-12