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cosx =1/3 => sinx = 2√2/3 or -2√2/3
case 1: sinx = 2√2/3
sin(x-π/2)/[ sin(π+x) -sin(3π/2-x) +1 ]
=-cosx/[ -sinx +cosx +1 ]
= -(1/3)/[ -2√2/3 + 1/3 +1 ]
=-1/( -2√2 + 4 )
=-( 2+√2)/4
case 2: sinx = -2√2/3
sin(x-π/2)/[ sin(π+x) -sin(3π/2-x) +1 ]
=-cosx/[ -sinx +cosx +1 ]
= -(1/3)/[ 2√2/3 + 1/3 +1 ]
=-1/( 2√2 + 4 )
=-( 2-√2)/4
case 1: sinx = 2√2/3
sin(x-π/2)/[ sin(π+x) -sin(3π/2-x) +1 ]
=-cosx/[ -sinx +cosx +1 ]
= -(1/3)/[ -2√2/3 + 1/3 +1 ]
=-1/( -2√2 + 4 )
=-( 2+√2)/4
case 2: sinx = -2√2/3
sin(x-π/2)/[ sin(π+x) -sin(3π/2-x) +1 ]
=-cosx/[ -sinx +cosx +1 ]
= -(1/3)/[ 2√2/3 + 1/3 +1 ]
=-1/( 2√2 + 4 )
=-( 2-√2)/4
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