Math 152 Trigonometry Rohl Fall 2017 Test Chapter 6 Test Thi

Math 152 Trigonometry- Rohl Fall 2017 Test: Chapter 6 Test This Question: 1 pt Timo 00:20:12 Submit Test 18 of 20 (17 complete) This Test:20 pts possible thera Solve the equation in the interval [0°,360) Use an algebraic method. utCH 9sin 20-4sin = 3 dLed Select the correct choice below and, if necessary, fil in the answer box to complete your choice elf (Simplify your answer. Round to the nearest tenth as needed. Use a comma to separate answers as needed. Do not include the degree symbol in your answer.) OA. The solution set is O B. The solution is the empty set Click to select and enter your answer(s) nnel

Solution

We have given 9sin²()-4sin()-3=0 in the interval [0⁰,360⁰)

9sin²()-4sin()-3=0

9sin²()-4sin()-3=0

by comparing ax^2+bx+c=0 we get a=9,b=-4,c=-3

sin()=[-b+/-sqrt(b^2-4ac)]/2a

sin()=[4+/-sqrt(16-(4*9*(-3)))]/18

sin()=[4+/-sqrt(16+108)]/18

sin()=[4+/-sqrt(124)]/18

sin()=[4+sqrt(124)]/18 , sin()=[4-sqrt(124)]/18

sin()=[4+2*sqrt(31)]/18 , sin()=[4-2*sqrt(31)]/18

sin()=[2+sqrt(31)]/9 , sin()=[2-sqrt(31)]/9

=arcsin((2+sqrt(31))/9),=arcsin((2-sqrt(31))/9)

=57.2313318⁰,=-23.3544569⁰

=-23.3544569⁰ excludes in the given interval

so =57.2313318⁰ is solution of the given equation

The solution set is {57.2}