How do you solve $Cos (pi/6 + x) + sin (pi/3 +x) = 0$ ?

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How do you solve $Cos (pi/6 + x) + sin (pi/3 +x) = 0$ ?

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Answer 1 · 2016-09-13T14:10:33

$pi/2, (3pi)/2$ for $(0, 2pi)$

Explanation:

Property of complementary arcs-->
$sin (pi/3 + x) = cos (pi/2 - (pi/3 + x)) = cos (pi/6 - x)$
Use trig identity:
$cos a + cos b = 2 cos ((a + b)/2).cos ((a - b)/2)$
The equation becomes:
$S = cos (pi/6 + x) + cos (pi/6 - x) = 0$
$a = (pi/6 + x)$ , and $b = (pi/6 - x)$
$(a + b)/2 = pi/6$ , and $(a - b)/2 = x$
Finally,
$S = 2cos (pi/6)cos x = 0$
cos x = 0 --> $x = pi/2$ and $x = (3pi)/2$
Answers for $(0, 2pi)$ :
$pi/2 ; (3pi)/2$
For general answers, add $2kpi$
Check with $x = (pi/2)$ :
$cos (pi/6 + x) = cos (pi/6 + pi/2) = cos ((2pi)/3) = - 1/2$
$sin (pi/3 + x) = sin (pi/3 + pi/2) = sin ((5pi)/6) = 1/2$
$S = -1/2 + 1/2 = 0$ . OK

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How do you solve $Cos (pi/6 + x) + sin (pi/3 +x) = 0$ ?

1 Answer

Explanation:

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How do you solve Cos (pi/6 + x) + sin (pi/3 +x) = 0Cos (pi/6 + x) + sin (pi/3 +x) = 0?

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How do you solve $Cos (pi/6 + x) + sin (pi/3 +x) = 0$ ?