Solve for x given {0 < x < 2}? $sin(2x) + cos(x) = 0$

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Answer 1 · 2018-06-12T16:20:53

$x in{pi/2, (3pi)/2, (7pi)/6, (11pi)/6}$

$sin2x+cosx=0--(1)$

$"for "{0 < x<2pi}$

$" now "sin2x=2sinxcosx$

$(1)rarr2sinxcosx+cosx=0$

$=>cosx(2sinx+1)=0$

$"either "cosx=0$

$=>x=pi/2, (3pi)/2$

$"or "2sinx+1=0$

$=>sinx=-1/2$

$:. x=(7pi)/6, (11pi)/6$

solution set

$x in{pi/2, (3pi)/2, (7pi)/6, (11pi)/6}$

Solve for x given {0 < x < 2}? sin(2x) + cos(x) = 0sin(2x) + cos(x) = 0