How do you solve $sin(2x) - cos(x) - 2sin(x) + 1 = 0$ ?

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Answer 1 · 2018-04-18T00:25:15

$sin(2x) - cos(x) - 2sin(x) + 1 = 0$

$=>2sin(x)cos(x) - cos(x) - 2sin(x) + 1 = 0$

$=>cos(x)(2sin(x) - 1) -( 2sin(x) - 1) = 0$

$=>(2sin(x) - 1)( cos(x) - 1) = 0$

When $2sin(x)-1=0$

$=>sin(x)=1/2=sin(pi/6)$

$=>x=npi+(-)^npi/6" where "n inZZ$

Again when

$cos(x)=1$

$=>x=2kpi" where " k inZZ$

How do you solve sin(2x) - cos(x) - 2sin(x) + 1 = 0sin(2x) - cos(x) - 2sin(x) + 1 = 0?