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Magazine Chapter 6: Problem 49
In Exercises \(37-54\), solve each of the trigonometric equations on \(0^{\circ} \leq \theta<360^{\circ}\) and express answers in degrees to two decimal places. $$ 15 \sin ^{2}(2 \theta)+\sin (2 \theta)-2=0 $$
Short Answer
Expert verified
The solutions are \( \theta \approx 9.74^\circ, 80.26^\circ, 101.79^\circ, \text{and} 331.21^\circ \).
Step by step solution
01
Substitute for Simplicity
Set a substitution such that let \( x = \sin(2\theta) \). The equation becomes \( 15x^2 + x - 2 = 0 \).
02
Identify Coefficients for the Quadratic Equation
The quadratic equation to solve is \( 15x^2 + x - 2 = 0 \). Here \( a = 15 \), \( b = 1 \), and \( c = -2 \).
03
Apply the Quadratic Formula
The roots of a quadratic equation \( ax^2 + bx + c = 0 \) are given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute \( a = 15 \), \( b = 1 \), \( c = -2 \) into the formula to find the values of \( x \).
04
Calculate the Discriminant
Calculate the discriminant, \( \Delta = b^2 - 4ac = 1^2 - 4(15)(-2) = 1 + 120 = 121 \).
05
Solve for x using the Roots Formula
Using the discriminant, \( x = \frac{-1 \pm \sqrt{121}}{30} \) which simplifies to \( x = \frac{-1 \pm 11}{30} \). This gives us \( x_1 = \frac{10}{30} = \frac{1}{3} \) and \( x_2 = \frac{-12}{30} = -\frac{2}{5} \).
06
Solve for \( \sin(2\theta) \) equals \( \frac{1}{3} \)
Notice \( \sin(2\theta) = \frac{1}{3} \). Use the inverse sine function to find \( 2\theta = \sin^{-1}\left(\frac{1}{3}\right) \). Compute \( 2\theta \approx 19.47^\circ \). Consider the possible angles, \( 2\theta \approx 180^\circ - 19.47^\circ = 160.53^\circ \).
07
Convert Solutions to \( \theta \) Values for \( \sin(2\theta) = \frac{1}{3} \)
Divide each \( 2\theta \) by 2 to find \( \theta \). Thus, \( \theta_1 \approx 9.74^\circ \) and \( \theta_2 \approx 80.26^\circ \).
08
Solve for \( \sin(2\theta) = -\frac{2}{5} \)
Similarly for \( \sin(2\theta) = -\frac{2}{5} \), we find \( 2\theta = \sin^{-1}\left(-\frac{2}{5}\right) \) which gives \( 2\theta \approx -23.58^\circ \). Also, calculate \( 2\theta = 180^\circ + 23.58^\circ = 203.58^\circ \).
09
Convert Solutions to \( \theta \) values for \( \sin(2\theta) = -\frac{2}{5} \)
Compute \( \theta \) by dividing the initial results by 2. The solution is \( \theta_3 = 331.21^\circ \) (adjust \( -23.58^\circ + 360^\circ \) within the range) and \( \theta_4 = 101.79^\circ \).
10
Collect All Solutions
Collect all the values of \( \theta \) found: \( 9.74^\circ, 80.26^\circ, 101.79^\circ, 331.21^\circ \).
11
Confirm the Angle Range
Verify all solutions \( 0^\circ \leq \theta < 360^\circ \) to ensure they satisfy the range requirement. All solutions are valid within this range.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Formula
To solve quadratic equations, the quadratic formula is a powerful tool. The quadratic formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula is used to find the values of \( x \) that satisfy the equation \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are coefficients from the equation. In this problem, we substituted \( x = \sin(2\theta) \) to express the trigonometric equation as a quadratic equation: \( 15x^2 + x - 2 = 0 \).
- Identify the coefficients as \( a = 15 \), \( b = 1 \), and \( c = -2 \).
- Calculate the discriminant: \( \Delta = b^2 - 4ac = 1^2 - 4 \times 15 \times (-2) = 121 \). This tells us the nature of the roots.
- Apply the quadratic formula: \( x = \frac{-1 \pm 11}{30} \). This results in the roots \( x_1 = \frac{1}{3} \) and \( x_2 = -\frac{2}{5} \).
Inverse Sine Function
Once we find the solutions \( x_1 \) and \( x_2 \) from the quadratic equation, we use the inverse sine function (\( \sin^{-1} \)) to find the angles that correspond to these values. The inverse sine function allows us to determine the angle \( \theta \) from a known sine value.
- If \( \sin(2\theta) = \frac{1}{3} \), we calculate \( 2\theta = \sin^{-1}(\frac{1}{3}) \approx 19.47^\circ \).
- Since sine is positive in the first and second quadrants, another possible angle is calculated by \( 180^\circ - 19.47^\circ = 160.53^\circ \).
- For \( \sin(2\theta) = -\frac{2}{5} \), we find \( 2\theta = \sin^{-1}(-\frac{2}{5}) \approx -23.58^\circ \).
- Adjust to the range by: \( 2\theta = 180^\circ + 23.58^\circ = 203.58^\circ \).
Angle Conversion
After finding values for \( 2\theta \), the next step involves angle conversion, specifically converting \( 2\theta \) into \( \theta \). This involves a simple yet important calculation.
- We calculated \( \theta \) by dividing each \( 2\theta \) value by 2.
- For \( 2\theta = 19.47^\circ \), divide by 2 to find \( \theta_1 = 9.74^\circ \).
- Similarly, for \( 2\theta = 160.53^\circ \), \( \theta_2 = 80.26^\circ \).
- For negative or out-of-range angles, like \( 2\theta = -23.58^\circ \), adjust to \( 336.42^\circ \) to find \( \theta = 336.42^\circ \).
- Finally, for \( 2\theta = 203.58^\circ \), find \( \theta = 101.79^\circ \).
