Solve each equation in x over the interval [0, 2π) and each equation in θ over the interval [0°, 360°). Give exact solutions.

√2 cos 2θ = -1

검증된 단계별 안내

Start with the given equation: \(\sqrt{2} \cos 2\theta = -1\).

Isolate the cosine term by dividing both sides by \(\sqrt{2}\): \(\cos 2\theta = \frac{-1}{\sqrt{2}}\).

Recognize that \(\cos 2\theta = -\frac{1}{\sqrt{2}}\) corresponds to specific angles where the cosine value is negative and equal to \(-\frac{1}{\sqrt{2}}\).

Find the general solutions for \(2\theta\) by identifying the reference angle whose cosine is \(\frac{1}{\sqrt{2}}\), which is \(45^\circ\) or \(\frac{\pi}{4}\) radians, and then determine the angles in the appropriate quadrants where cosine is negative (quadrants II and III).

Divide the solutions for \(2\theta\) by 2 to solve for \(\theta\), and then restrict the solutions to the given interval \([0^\circ, 360^\circ)\) or \([0, 2\pi)\) depending on the unit.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Double-Angle Trigonometric Functions

The equation involves cos 2θ, a double-angle function representing the cosine of twice the angle θ. Understanding how to manipulate and solve equations with double angles is essential, as it changes the period and affects the solution set within the given interval.

Solving Trigonometric Equations

Solving trigonometric equations requires isolating the trigonometric function and finding all angles within the specified interval that satisfy the equation. This often involves using inverse trigonometric functions and considering the periodic nature of sine and cosine.

Interval and Exact Solutions in Radians and Degrees

The problem specifies solutions over [0, 2π) for x and [0°, 360°) for θ, requiring careful attention to the domain. Providing exact solutions means expressing answers in terms of π or degrees rather than decimal approximations, ensuring precision and completeness.

Solve each equation in x over the interval [0, 2π) and each equation in θ over the interval [0°, 360°). Give exact solutions.√2 cos 2θ = -1