Math  /  Trigonometry

QuestionQuestion 3 (2 points) Solve the following equation for 0θ2π0 \leq \theta \leq 2 \pi. Select all correct solutions. 2sin2θ1=02 \sin ^{2} \theta-1=0
Select 4 correct answer(s)

Studdy Solution
Determine all angles θ\theta within the interval 0θ2π0 \leq \theta \leq 2\pi that satisfy sinθ=±22\sin \theta = \pm \frac{\sqrt{2}}{2}.
The angles where sinθ=22\sin \theta = \frac{\sqrt{2}}{2} are:
θ=π4,3π4 \theta = \frac{\pi}{4}, \frac{3\pi}{4}
The angles where sinθ=22\sin \theta = -\frac{\sqrt{2}}{2} are:
θ=5π4,7π4 \theta = \frac{5\pi}{4}, \frac{7\pi}{4}
The correct solutions for θ\theta are:
π4,3π4,5π4,7π4 \boxed{\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}}

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