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

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Answer 1 · 2018-06-12T16:20:53

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

$sin 2 x + cos x = 0 - - (1)$ $sin2x+cosx=0--(1)$

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

$now sin 2 x = 2 sin x cos x$ $" now "sin2x=2sinxcosx$

$(1) \to 2 sin x cos x + cos x = 0$ $(1)rarr2sinxcosx+cosx=0$

$\Rightarrow cos x (2 sin x + 1) = 0$ $=>cosx(2sinx+1)=0$

$either cos x = 0$ $"either "cosx=0$

$\Rightarrow x = \frac{π}{2}, \frac{3 π}{2}$ $=>x=pi/2, (3pi)/2$

$or 2 sin x + 1 = 0$ $"or "2sinx+1=0$

$\Rightarrow sin x = - \frac{1}{2}$ $=>sinx=-1/2$

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

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)=0sin(2x) + cos(x) = 0