Solve the equation #sin^2x-1/2 sinx-1/2=0# where #0lexle2pi# ?

2 Answers
Lucy
Apr 14, 2018

#x=pi/2, (7pi)/6, (11pi)/6#

Explanation:

#(sinx)^2-1/2sinx-1/2=0#

#2(sinx)^2-sinx-1=0#

#(2sinx+1)(sinx-1)=0#

#2sinx+1=0# or #sinx-1=0#

#sinx=-1/2#

#x=(7pi)/6, (11pi)/6#

#sinx=1#

#x=pi/2#

WADU HEK
Apr 14, 2018

x = pi/2

Explanation:

1/ #sin^2x# - #1/2 sinx - 1/2#
2/ Factor it = #(sinx + 1/2)(sinx-1) #
3/ sinx = #-1/2# ; sinx = 1
4/ #x = sin^-1(1/2) ; x = sin^-1 (1)#
5/# x = pi/6 ; x = pi/2#
6/ Check both answer with your calculator and see which one works

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