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Question

#sin2x-2cosx+0.25=0#

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Answer 1 · 2018-01-24T12:10:22

#sin(2x)=2sinx cosx# then

#sin2x-2cosx+0.25=2sinx cosx-2cosx+1/4=0#

then

#cos x = 1/(8(sinx-1))# but

#sin^2x+cos^2x = 1# so

#sin^2x+1/(64(sinx-1)^2)=1# and now solving for #sin x# we have the real solutions as

#sinx = {(-0.998041),(0.794264):}#

The next steps are left to the reader.