Exercises 39–52 involve trigonometric equations quadratic in form. Solve each equation on the interval [0, 2𝝅). 2 cos² x + 3 cos x + 1 = 0

검증된 단계별 안내

Recognize that the equation is quadratic in form with respect to \( \cos x \). Let \( u = \cos x \), so the equation becomes \( 2u^{2} + 3u + 1 = 0 \).

Solve the quadratic equation \( 2u^{2} + 3u + 1 = 0 \) using the quadratic formula: \( u = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \), where \( a=2 \), \( b=3 \), and \( c=1 \).

Calculate the discriminant \( \Delta = b^{2} - 4ac = 3^{2} - 4 \times 2 \times 1 \) and then find the two possible values for \( u = \cos x \).

For each value of \( \cos x \) found, determine the corresponding values of \( x \) in the interval \( [0, 2\pi) \) by using the inverse cosine function: \( x = \arccos(u) \) and considering the symmetry of cosine in the unit circle.

Write down all solutions \( x \) in \( [0, 2\pi) \) that satisfy the original equation, ensuring to check if any values of \( u \) are outside the valid range for cosine (i.e., \( -1 \leq u \leq 1 \)) and discard those if necessary.

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주요 개념

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Quadratic Form in Trigonometric Equations

Some trigonometric equations can be rewritten as quadratic equations by substituting a trigonometric function (e.g., cos x) with a variable. This allows the use of algebraic methods like factoring or the quadratic formula to find solutions for the trigonometric function.

Solving Quadratic Equations

Solving quadratic equations involves finding values of the variable that satisfy the equation, typically by factoring, completing the square, or using the quadratic formula. These solutions correspond to values of the trigonometric function in the original equation.

Finding Angles from Trigonometric Values on a Given Interval

After determining the values of the trigonometric function, you find all angles x within the specified interval [0, 2π) that satisfy these values. This requires understanding the unit circle and the periodic nature of trigonometric functions to identify all valid solutions.