Practice Problems for 6.2 and 6.3 (Trigonometric Equations)

THREE: Solve the following trigonometric equation: Find all solutions where 0 ≤ x ≤ 2π tan²(x) - 1 = 0 *answer in radians*

1. Add one to both sides 2. Take the square root of both sides. The complete solution set is the positive and negative portions of the solution: tanx = 1 and tanx = -1 3. Using the inverse tangent function, we know that a 45° reference angle has a function value of 1 for tanx = 1 and -1 for tanx = -1. 4. Using the reference angle, solve x for the positive and negative portions. 5. Solving for x using the positive portion. Tangent is positive in Quadrant I and Quadrant III. In Quadrant I θ = θ', so θ = π/4. In Quadrant II θ = π + θ', so θ = π + π/4 = 5π/4. 6. Solving for x using the negative portion. Tangent is negative in Quadrant II and Quadrant IV. In Quadrant II θ = π - θ', so θ = π - π/4 = 3π/4. In Quadrant IV θ = 2π - θ', so θ = 2π - π/4 = 7π/4 7. The solutions are π/4, 3π/4, 5π/4, and 7π/4.

SEVEN: Solve the equations on the interval [0, 2π) cosxtanx - cosx = 0

1. Factor out cosx and use the zero-factor property 2. For cosx take the arccos of both sides to extract x from inside cos and to find the reference angle to solve for x: arccos(cosx) = arccos (0). An angle with the function value of 0 is π/2 and 3π/2. 3. To solve for the other solution add one to both sides of the equation and take the arccos of both sides to find the reference angle. Tangent is positive in Quadrant I and III. In Quadrant I θ = θ', so θ = π/4. In Quadrant III θ = π + θ', so θ = π + π/4 = 5π/4 4. The solutions are π/2, 3π/2, π/4, 5π/4

NINE: Solve 5cos(2x) = 5sin²(x) + 4 for all solutions 0 ≤ x < 2π x= _____________ *Give your answers accurate to at least 2 decimal places

1. Subtract 5sin²x + 4 to both sides 2. Use the double-angle identity cos(2x) = 1-2sin²(x) 3. Distribute 4. Subtract 5sin²x from 10sin²x 5. Subtract 4 from 5 6. Subtract 1 from both sides and divide by -15 7. sin²x = 1/15 8. sinx = +/- √1/15. take the square root of both sides 9. sinx = + √15/15 and sinx = -√15/15 √1 = 1 and the denominator was rationalized. When taking the square root of both sides solve for the positive and negative portion PART 2 10. Take arcsin of both equations to get x out of sin and to find the reference angle to solve for x. 11. x = arcsin √15 and x = arcsin = - √15 12. x= .26, -.26 13. Sine is positive in Quadrant I and II. In Quadrant one θ = θ', so θ = .26. In quadrant II, θ = π - θ ', so θ = π - .26 = 2.88. The other solutions for -.26 yield 2.88 in Quadrant 3 and and 6.54 in Q IV, which is outside the interval. 14. To find all solutions, add the period of arcsin to each solution. All solutions within the interval [0, 2π] should be listed: .26 + π = 3.40 2.88 + π = 6.02 X = .26, 2.88, 3.40, and 6.02

TEN: Find all solutions on the interval [0, 2π) to the following equation. Answer should be in exact form cos(2x) = 7sin(x) - 3 x = _________..

1. Subtract 7sinx and add 3 from/to both sides 2. Use double-angle identity cos(2x) = 1-2sin²x 3. Add 1 and 4 on the LHS = -2sin²x - 7sinx + 4 = 0 4. -2sin²x + sinx - 8sinx + 4 = 0: Reorder terms for a polynomial of the form ax²+bx+c, rewrite the middle term as a sum of two terms whose product is a • c = -2 • 4 = -8 and whose sum is b = -7 5. Factor out the GCF of each group: sinx(-2sinx + 1) + 4(-2sinx + 1) = 0 6. Factor the polynomial by factoring out the GCF: (-2sinx + 1)(sinx+4) = 0 7. Use the zero-factor property: -2sinx + 1 = 0 and -sinx + 4 = 0 8. -2sinx + 1 = 0 subtract 1 from both sides and divide both sides by -2, -1/-2 = 1/2. 9. -sinx + 4 = 0 add 4 to both sides 10. Use arcsin to find the reference angle for both. Sine is positive in Quadrant I and II. In quadrant one θ = θ'. In quadrant II θ = π - θ'. So in quadrant one θ = π/6 and in quadrant II θ = 5π/6. 11. arcsin (-4) is outside of the interval for arcsin so DNE.

SIX: Solve the equations on the interval [0, 2π) 1 + √3tanθ = 2

1. Subtract one from both sides. 2. Divide both sides by the √3 and rationalize the denominator. 3. Take the inverse tangent of both sides to extract x from inside tangent on the LHS and to solve for x. The inverse function yields a reference angle. 4. θ = arctan √3/3, which yields a reference angle of π/6. Use the reference angle to solve for θ. 5. Tangent is positive in Quadrant I and III. In Quadrant one θ = θ', so θ = π/6. In Quadrant III θ = π + θ', so θ = π + π/6 = 7π/6. 6. X = π/6, 7π/6

FOUR: Solve the following trigonometric equation: Find all solutions where 0 ≤ x ≤ 2π √2cos(x) + 1 = 0 *answer in radians*

1. Subtract one from both sides. 2. Divide both sides by √2 3. Rationalize the denominator 4. Take inverse cosine of both sides to extract x from inside cosine on the LHS and to find the reference angle. 5. The reference angle is π/4. Use the reference angle to find the value of x. 6. Cosine is negative in Quadrant II and III. In Q II θ = π - θ', so θ = π - π/4 = 3π/4. In Quadrant III θ = π + θ', so θ = π + π/4 = 5π/4.

Solve the following trigonometric equation: Find all solutions where 0 ≤ x ≤ 2π cos²(x) + 1 = 4 *answer in radians*

1. Subtract one from both sides. 2. Take the square root of both sides to get rid of the exponent on the LHS. The complete solution set is the positive and negative portions of the solution: cosx = √3 and cosx = -√3. 3. We would then take the inverse cosine of each side to extract x from cos and to find the value of x. The range of arccos is −1 ≤ y ≤1. Since √3 does not fall in this range, there is no solution. A calculator will give an error message. No solution

EIGHT: Find all solutions on the interval 0 ≤ x < 2π 2tan(x)sin(x) = √3tan(x) *Give answers in exact form, separated by commas*

1. Subtract √3tanx from both sides. 2. Factor out tanx 3. Use the zero-factor property 4. sin(x) - √3 = 0: Add √3 to both sides and divide by 2 on both sides. Then take arcsin of both sides to extract x from inside of sin and to find the reference angle with a function value of √3/2 to use it to solve for x. Sine is positive in Quadrant I and III. In Quadrant I θ = θ', so θ = π/3. In Quadrant II θ = π - θ', so θ = π - π/3 = 2π/3. 5. Tan(x) = 0: take the arctan of both sides. The function value of tangent is 0 when θ = 0° and π. 6. X = 0°, π, π/3, 2π/3.

TWO: Solve the following trigonometric equation: Find all solutions where 0 ≤ x ≤ 2π sin²(x) = 1/2 *answer in radians*

1. Take the square root of both sides to eliminate the exponent on the LHS. = sinx = +/- √1/2 2. sinx = +/- 1/√2. √1 = 1 3. sinx = +/- √2/2 rationalize denominator 4. sinx = √2/2 and sinx = - √2/2. The complete solution is the result of both the positive and the negative portions of the solution. 5. arcsin(sinx) = arcsin (√2/2). and arcsin(sinx) = arcsin (- √2/2). Solve for x by taking the inverse sine function of both sides to extract x from inside sine on the LHS. 6. The inverse sine function finds the reference angle which is π/4 for the positive portion and -π/4 for the negative portion. 7. Using the reference angles found, solve for x. Sine is positive in Quadrant I and II. In Quadrant I θ = θ' so our first solution is π/4. To find the second positive solution in Quadrant II, θ = π - θ', so θ = π - π/4 = 3π/4. 8. To find the other two solutions for the negative portion, start with Quadrant III. In Quadrant III θ = π + θ', so θ = π + π/4 = 5π/4. In Quadrant IV θ = 2π - θ', so θ = 2π - π/4 = 7π/4. So the solutions are π/4, 3π/4, 5π/4, and 7π/4.

ELEVEN: Find all solutions to the given equations. Use a graphing utility to graph each side of the given equation to check your solutions. (Write each solution in terms of an angle on [0,2π). If an equation has multiple solutions, order them from the smallest angle to the largest.) (a) cos(x)sin(x) = 1/2 (b) cos(2x) + 3 = 5cos(x)

PART A 1. Clear fractions by multiplying all terms by 2 2. Use double angle identity for sin(2x) = 2sin(x)cos(x) --same as 2cos(x)sin(x) 3. Take the inverse sine function of both sides to extract x from within sine on the LHS and to find the reference angle to solve for x. 4. Use a calculator or unit circle to find arcsin(1) 5. Divide each term by 2 6. x = π/4 + πK. add the period of cosine/2, which is 2π/2 = π (this was written wrong on the whiteboard--its not the period of arcsin) PART B 1. Subtract 5cos(x) from both sides. 2. Use the double-angle identity cos(2x) = 2cos²x - 1 3. Combine 1 and 3 4. Rewrite in ax² + bx + c. Rewrite the middle term as a sum of two terms whose product is a • c = 2 • 2 = 4 and whose sum is b = -5 5. Group the terms and factor out the GCF of each group 6. Factor the polynomial by factoring out the GCF 7. Use the zero-factor property 8. Isolate cos(x) on the LHS of each equation and take the arccos of both equations to isolate x outside of cosine to find the reference angle and solve for x. 9. Cosine is positive in Quadrant I and IV. IN Quadrant one θ = π/3 and in Quadrant IV θ = 5π/3 10. X = π/3 + 2πK and 5π/3 + 2πK. with K being any integer. to find all solutions the period of cosine is added to each solution.

ONE: Problem worked out on whiteboard

The problem involves one trigonometric function.