How do you solve? 5cos2x+3sinx=4
1 Answer
Dec 18, 2017
See the answer below....
Explanation:
#5cos2x+3sinx=4#
#=>5(1-2sin^2x)+3sinx=4#
#=>5-10sin^2x+3sinx=4#
#=>10sin^2x-3sinx-1=0#
#=>10sin^2x-5sinx+2sinx-1=0#
#=>5sinx(2sinx-1)+1(2sinx-1)=0#
#=>(5sinx+1)(2sinx-1)=0# Now,either
#" "sinx=-1/5" or. "sinx=1/2# Now,you can determine the value of
#x# .Hope it helps.....thank you...
