How to solve this equation? $2 cos x + 1 = 0; x \in [0, 3 π)$ $2cosx+1=0;x in[0,3pi)$ ;

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Answer 1 · 2017-04-05T17:28:20

$\frac{2 π}{3}, \frac{4 π}{3}, \frac{8 π}{3}$ $(2pi)/3, (4pi)/3, (8pi)/3$

$2 cos x + 1 = 0$ $2cosx+1=0$

$cos x = - \frac{1}{2}$ $cosx= -1/2$

Now we know from this handy "CAST" nmemonic that cos is negative in Q2+Q3:

Rembrandt

We can even draw the triangles:

Picasso

So in Q1: $cos x = \frac{1}{2} \Rightarrow x = \frac{π}{3}$ $cos x = 1/2 implies x = pi/3$

The corresponding angle in Q2 is $π - \frac{π}{3} = \frac{2 π}{3}$ $pi - pi/3 = (2pi)/3$

The corresponding angle in Q3 is $π + \frac{π}{3} = \frac{4 π}{3}$ $pi + pi/3 = (4pi)/3$

Now because it is $x \in [0, 3 π)$ $x in[0,3pi)$ , we are not done yet.

After spinning round a full $2 π$ $2 pi$ , we have another $π$ $pi$ to go which takes us back into Q2.

The second corresponding angle in Q2 is $2 π + (π - \frac{π}{3}) = \frac{8 π}{3}$ $2 pi + (pi - pi/3) = (8pi)/3$

How to solve this equation? 2cosx+1=0;x in[0,3pi)2cosx+1=0;x∈[0,3π)2cosx+1=0;x in[0,3pi);