Find all solutions for the following equations sin(2x)+sin(π/2-x)=0 ?

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Find all solutions for the following equations sin(2x)+sin(π/2-x)=0 ?

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Answer 1 · 2018-05-14T23:46:18

$\frac{π}{2}; \frac{3 π}{2}; \frac{7 π}{6}; \frac{11 π}{6}$ $pi/2; (3pi)/2 ; (7pi)/6; (11pi)/6$ $(+ 2 k π)$ $(+ 2kpi)$

Explanation:

$sin 2 x + sin (\frac{π}{2} - x) = 0$ $sin 2x + sin (pi/2 - x) = 0$
Note. sin 2x = 2sin x.cos x
$sin (\frac{π}{2} - x) = cos x$ $sin (pi/2 - x) = cos x$
Therefor,
$2 sin x . cos x + cos x = 0$ $2sin x.cos x + cos x = 0$
$cos x (2 sin x + 1) = 0$ $cos x(2sin x + 1) = 0$
Either factor should be zero -->
a. $cos x = 0$ $cos x = 0$
Unit circle gives -->
$x = \frac{π}{2} + 2 k π$ $x = pi/2 + 2kpi$ , and
$x = \frac{3 π}{2} + 2 k π$ $x = (3pi)/2 + 2kpi$
b. $2 sin x + 1 = 0$ $2sin x + 1 = 0$ --> $sin x = - \frac{1}{2}$ $sin x = - 1/2$
Unit circle gives -->
$x = \frac{7 π}{6} + 2 k π$ $x = (7pi)/6 + 2kpi$ , and
$x = \frac{11 π}{6} + 2 k π$ $x = (11pi)/6 + 2kpi$

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Find all solutions for the following equations sin(2x)+sin(π/2-x)=0 ?

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