What is the number of degrees in the measure of angle #x# in the interval #0°<=x<=270°# that satisfies the equation #2sin^2(x) - 5sinx - 3=0#?

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Answer 1 · 2016-03-14T05:51:10

#x={150^o,240^o}#

We can solve #2sin^2x-5sinx-3=0# for #sinx# using quadratic formula

Hence #sinx=(-(-5)+-sqrt((-5)^2-4xx2xx(-3)))/(2xx2)#

= #(5+-sqrt(25+24))/(2xx2)#

= #(5+-7)/4# or

= #3 or -1/2#

As #sinx# has to be within the range #[-1,1]#

#sinx=-1/2#

Within the range #0^o<=x<=270^o#, #sinx=-1/2# for

#x={150^o,240^o}#