A basic trigonometric equation asks what values of the trig function have a specific value.
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1 Lecture 3A: Solving Basic Trig Equations A basic trigonometric equation asks what values of the trig function have a specific value. The equation sinθ = 1 asks for what vales of θ is the equation true. We know that sin π = 1 follows that θ = π is a solution to the equation. It in NOT the only solution. We know that sin 5π and sin 13π = 1 17π and sin = 1 so θ = π,5π,13π and 17π so it = 1 are also solutions. In fact, there are an infinite number of solutions to any basic trig equation. The first part of this section reduces the number of solutions by limiting the domain for θ to the values on the interval where θ is on the interval [ 0, π ). The last part of the section requires that all the possible solutions are listed. Problem 1 Solve the equation sinθ = 1 for θ on the interval [ 0, π ) sin( θ ) equals 1 at θ = π and θ = 5π This means thatθ = π and θ = 5π are the solutions to the equation sin( θ ) = 1 on [0,π) Problem Solve the equation sinθ = 1 for θ on the interval [ 0, π ) sin( θ ) equals 1 at θ = 7π and θ = 11π for θ on the interval [ 0, π ) This means that θ = 7π and θ = 11π are are the solutions to the equation sin ( θ ) = 1 on [0,π) Problem 3 Solve the equation sin( θ ) = 1 for θ on the interval [ 0, π ) sin( θ ) equals 1 at θ = 3π on the interval [ 0, π ) This means that θ = 3π is the only solution to the equation sin( θ ) = 1 on [0,π) Math 370 Section 3A Page Eitel
2 Solution Techniques There are several ways to find the solutions to a basic trig equation. One common way is to have the unit circle memorized. A student could then think of the angles were the function has the specified values and write down those answers. That was the method used to find the answers to the problems shown above. Some students require a more defined method. That requires the use of the first quadrant reference angles and the chart that shows which quadrants the trig functions are positive or negative. Solve sin ( θ ) = 1 Example 1 π We know sin = y = 1 The reference angle that solves the equation for a POSITIVE value is π. Find the quadrants where the function has the required sign. sin θ ( ) is positive in the 1st and nd quadrants. The solutions for θ are π in the 1st and nd quadrants θ = π and 5π Math 370 Section 3A Page 015 Eitel
3 Example Solve cos ( θ ) = 3 π We know cos = x = 3 The reference angle that solves the equation for a POSITIVE value is π. Find the quadrants where the function has the required sign. cos θ ( ) is negitive in the nd and 3rd quadrants. The solutions for θ are π in the nd and 3rd quadrants. θ = 5π and 7π Math 370 Section 3A Page Eitel
4 Example 3 Solve tan ( θ ) = 1 π We know tan = y x = 1 The reference angle that solves the equation for a POSITIVE value is π. Find the quadrants where the function has the required sign. tan θ ( ) is negitive in the nd and 4th quadrants. The solutions for θ are π 4 in the nd and 4th quadrants. θ = 3π 4 and 7π 4 Math 370 Section 3A Page Eitel
5 Example 4 Solve sec ( θ ) = 3 3 π We know sec = 1 x = 3 3 The reference angle that solves the equation for a POSITIVE value is π. Find the quadrants where the function has the required sign. sec θ ( ) is positive in the 1st and 4th quadrants. The solutions for θ are π in the 1st and 4th quadrants. θ = π and 11π Math 370 Section 3A Page Eitel
6 Example 4 Quadrangles Solve sin ( θ ) = 0 Look at the quadrangle chart and find where sin ( θ ) = 0 sin ( θ ) = y and y = 0 when θ = 0 and θ = π Example 5 Quadrangles Solve cos ( θ ) = 0 Look at the quadrangle chart and find where cos ( θ ) = 0 cos ( θ ) = x and x = 0 when θ = π and θ = 3π Example Quadrangles where the is only 1 solution for θ Solve cos ( θ ) = 1 Look at the quadrangle chart and find where cos ( θ ) = 1 cos ( θ ) = x and x = 1 only when θ = π Example 7 Quadrangles where the is only 1 solution for θ Solve sin ( θ ) = 1 Look at the quadrangle chart and find where sin ( θ ) = 1 sin ( θ ) = y and y = 1 only when θ = 3π Math 370 Section 3A Page 015 Eitel
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