How do you solve the equation #sintheta=0.91# for #0<theta<2pi#?

2 Answers
May 25, 2018

#theta ~~ 65.5^@ " and " ~~114.5^@#

Explanation:

Given: #sin theta = 0.91# for #0 < theta < 2pi#

The sine is positive in two quadrants: quadrant 1 & 2

#theta = sin^-1 (0.91) ~~ 65.5^@#

To find #theta# in the 2nd quadrant, subtract this angle from #180^@#:

#theta = 180^@ - 65.5^@ = 114.5^@#

May 25, 2018

#theta=1.14" or "theta=2#

Explanation:

#"since "sintheta>0" then "theta" is in first or second"#
#"quadrant"#

#theta=sin^-1(0.91)=1.14" radians"larrcolor(red)"first quadrant"#

#"or "theta=(pi-1.14)=2larrcolor(red)"in second quadrant"#

#"solutions are "theta=1.14" or "theta=2to(0,2pi)#