Solving Trig Equations with Calculators, Part II — Question 1

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Question 1

Solve the equation 2cos⁡(x)+1=02\cos(x) + 1 = 0 for all x∈[0,2π]x \in [0, 2\pi], and round your answers to two decimal places if necessary.

Original worksheet page 1: question and worked solution for 1-6-001
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Question 1 - Solution

Step 1: Isolate the cosine term

2cos⁡(x)+1=0⇒cos⁡(x)=−122\cos(x) + 1 = 0 \Rightarrow \cos(x) = -\frac{1}{2}

Step 2: Solve the equation using inverse cosine

cos⁡−1(12)=π3⇒Reference angle=π3\cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \Rightarrow \text{Reference angle} = \frac{\pi}{3}

Since cosine is negative, the solutions are in Quadrants II and III:

- Quadrant II: x=π−π3=2π3x = \pi - \frac{\pi}{3} = \frac{2\pi}{3} - Quadrant III: x=π+π3=4π3x = \pi + \frac{\pi}{3} = \frac{4\pi}{3}

x=2π3,4π3\boxed{x = \frac{2\pi}{3},\ \frac{4\pi}{3}}

Original worksheet page 2: question and worked solution for 1-6-001

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