How do you solve sin(x)-cos(x/3)=0?

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How do you solve sin(x)-cos(x/3)=0?

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Answer 1 · 2018-05-17T22:05:25

$x = \frac{3 π}{8}$ $x=(3pi)/8$

Explanation:

.

$sin x - cos (\frac{x}{3}) = 0$ $sinx-cos(x/3)=0$

$sin x = cos (\frac{x}{3})$ $sinx=cos(x/3)$

We know:

$sin (\frac{π}{2} - θ) = cos θ$ $sin(pi/2-theta)=costheta$

Therefore,

$sin (\frac{π}{2} - \frac{x}{3}) = cos (\frac{x}{3})$ $sin(pi/2-x/3)=cos(x/3)$

$x = \frac{π}{2} - \frac{x}{3}$ $x=pi/2-x/3$

$x + \frac{x}{3} = \frac{π}{2}$ $x+x/3=pi/2$

$6 x + 2 x = 3 π$ $6x+2x=3pi$

$8 x = 3 π$ $8x=3pi$

$x = \frac{3 π}{8}$ $x=(3pi)/8$

Answer 2 · 2018-05-17T22:22:41

$x = \frac{3 π}{8} + \frac{3 k π}{2}$ $x = (3pi)/8 + (3kpi)/2$
$x = \frac{3 π}{4} + 3 k π$ $x = (3pi)/4 + 3kpi$

Explanation:

$sin x = cos (\frac{x}{3})$ $sin x = cos (x/3)$
$cos (\frac{π}{2} - x) = cos (\frac{x}{3})$ $cos (pi/2 - x) = cos (x/3)$
Unit circle and property of cos function -->
$\frac{π}{2} - x = \pm \frac{x}{3}$ $pi/2 - x = +- x/3$
a. $\frac{π}{2} - x = \frac{x}{3}$ $pi/2 - x = x/3$
$x + \frac{x}{3} = \frac{π}{2}$ $x + x/3 = pi/2$
$\frac{4 x}{3} = \frac{π}{2} + 2 k π$ $(4x)/3 = pi/2 + 2kpi$
$x = \frac{3 π}{8} + \frac{3 k π}{2}$ $x = (3pi)/8 + (3kpi)/2$
b. $\frac{π}{2} - x = - \frac{x}{3}$ $pi/2 - x = - x/3$
$x - \frac{x}{3} = \frac{π}{2}$ $x - x/3 = pi/2$
$(2 \frac{x}{3}) = \frac{π}{2} + 2 k π$ $(2x/3) = pi /2 + 2kpi$
$x = \frac{3 π}{4} + 3 k π$ $x = (3pi)/4 + 3kpi$

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How do you solve sin(x)-cos(x/3)=0?

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