Solve the equation over the interval [0, 2π)?
Solve the equation over the interval [0, 2π). Round to three decimal places. (Enter your answers in radians. Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.)
11 sin^2(θ) = 9
Solve the equation over the interval [0, 2π). Round to three decimal places. (Enter your answers in radians. Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.)
11 sin^2(θ) = 9
2 Answers
Explanation:
Explanation:
#"all answers are given in radians to 3 decimal places"#
#"isolate "sin^2theta" by dividing both sides by 11"#
#rArrsin^2theta=9/11#
#rArrsintheta=+-sqrt(9/11)larrcolor(blue)"note plus or minus"#
#sintheta=sqrt(9/11)totheta" in first/second quadrant"#
#sintheta=-sqrt(9/11)totheta" in third/fourth quadrants"#
#rArrtheta=sin^-1(sqrt(9/11))=1.130larrcolor(blue)"first quad"#
#rArrtheta=(pi-1.130)=2.012larrcolor(blue)"second quad"#
#rArrtheta=(pi+1.130)=4.272larrcolor(blue)"third quad"#
#rArrtheta=(2pi-1.130)=5.153larrcolor(blue)"fourth quad"#
#rArrtheta=1.130,2.012,4.272,5.153#