How do you solve #12sinx-7sqrtsinx+1=0# in the interval [0,360]?
1 Answer
Dec 20, 2016
Start by isolating the square root.
#12sinx +1 = 7sqrt(sinx)#
#(12sinx + 1)^2 = (7sqrt(sinx))^2#
#144sin^2x + 24sinx + 1 = 49sinx#
#144sin^2x - 25sinx + 1 = 0#
We let
#144t^2 - 25t + 1 = 0#
Solve by factoring:
#144t^2 - 16t - 9t + 1 = 0#
#16t(9t - 1) - 1(9t - 1) = 0#
#(16t - 1)(9t - 1) = 0#
#t = 1/16 and 1/9#
#sinx = 1/16 and sinx = 1/9#
#x = arcsin(1/16), 180˚ - arcsin(1/16), arcsin(1/9) and 180˚ - arcsin(1/9)#
Make sure your calculator is in degree mode when approximating these values.
After checking your answers for extraneous solutions, you should get that all solutions work.
Hopefully this helps!
