Question

1. Solve for $0 \leq \theta \leq 2\pi$: \newline $\sin\theta = \frac{1}{2}$ \newline \newline 2. Solve for $-\pi \leq \theta \leq \pi$: \newline $\cos\theta = -\frac{\sqrt{2}}{2}$ \newline \newline 3. Solve for $\theta \in \mathbb{R}$: \newline $\tan\theta = 1$ \newline \newline 4. Solve for $\theta \in \mathbb{R}$: \newline $2\cos^2\theta - 7\cos\theta + 3 = 0$

Name: 1. Solve for 0 ≤θ≤ 2π: sinθ = (1)/(2) 2. Solve for -π≤θ≤π: cosθ = -(√(2))/(2) 3. Solve for θ∈ℝ: tanθ = 1 4. Solve for θ∈ℝ: 2cos^2θ - 7cosθ + 3 = 0
Uploaded: 2019-08-08T07:53:51-08:00
Duration: 41 s
Channel: Clarissa Noh
Description: 1. Solve for 0 ≤θ≤ 2π: sinθ = (1)/(2) 2. Solve for -π≤θ≤π: cosθ = -(√(2))/(2) 3. Solve for θ∈ℝ: tanθ = 1 4. Solve for θ∈ℝ: 2cos^2θ - 7cosθ + 3 = 0

          1. Solve for $0 \leq \theta \leq 2\pi$: \newline $\sin\theta = \frac{1}{2}$ \newline \newline 2. Solve for $-\pi \leq \theta \leq \pi$: \newline $\cos\theta = -\frac{\sqrt{2}}{2}$ \newline \newline 3. Solve for $\theta \in \mathbb{R}$: \newline $\tan\theta = 1$ \newline \newline 4. Solve for $\theta \in \mathbb{R}$: \newline $2\cos^2\theta - 7\cos\theta + 3 = 0$

1. Solve for 0 ≤θ≤ 2π: sinθ = (1)/(2) 2. Solve for -π≤θ≤π: cosθ = -(√(2))/(2) 3. Solve for θ∈ℝ: tanθ = 1 4. Solve for θ∈ℝ: 2cos^2θ - 7cosθ + 3 = 0

Added by Alyssa D.

Precalculus with Limits

Ron Larson 2nd Edition

Instant Answer

Solved by Expert Clarissa Noh

08/08/2019

Step 1

The reference angle for sinθ = (1)/(2) is π/6. So, the solutions are θ = π/6 and θ = 5π/6. Show more…

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Solve for 0 <= θ <= 2π: sinθ = (1)/(2) Solve for θ in R: tanθ = 1 Solve for -π <= θ <= π: cosθ = -(√2)/(2) Solve for θ in R: 2cos^2θ - 7cosθ + 3 = 0 1. Solve for 0 <= 6 <= 2x = us 2. Solve for fer: ccse = - 3. Solve for 1nS = 1 4. Solve for R: 2cos7cas6 + 3 = 0