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

2 Answers
Feb 18, 2018

#bb1.130color(white)(88) bb2.011color(white)(88)bb4.272color(white)(88)bb5.153#

Explanation:

#11sin^2(theta)=9#

#sin^2(theta)=9/11#

#sin(theta)=+-sqrt(9/11)#

#theta=arcsin(sintheta)=arcsin(sqrt(9/11))=1.1303, pi-1.1303#

#theta=arcsin(sintheta)=arcsin(-sqrt(9/11))=pi+1.1303, 2pi-1.1303#

#bb1.130color(white)(88) bb2.011color(white)(88)bb4.272color(white)(88)bb5.153#

Feb 18, 2018

#"see 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#