5sinx=5sqrt3cosx 0<x<2pi What are all possible solutions?

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5sinx=5sqrt3cosx 0<x<2pi What are all possible solutions?

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Answer 1 · 2015-10-25T03:10:46

solve $5 sin x = 5 \sqrt{3} cos x$ $5sin x = 5sqrt3cos x$

Ans: $\frac{π}{3} and \frac{4 π}{3}$ $pi/3 and (4pi)/3$

Explanation:

There are two ways.
1. First way. Divide both side by 5cos x (condition $cos x \neq 0, or x \neq \frac{π}{2} and x \neq 3 \frac{π}{2}$ $cos x != 0, or x != pi/2 and x !=3pi/2$ )
$tan x = \sqrt{3}$ $tan x = sqrt3$ --> $x = \frac{π}{3}$ $x = pi/3$ and $x = \frac{π}{3} + π = \frac{4 π}{3}$ $x = pi/3 + pi = (4pi)/3$
2. Second way. Simplify both sides by 5
$sin x - \sqrt{3} cos x = 0$ $sin x - sqrt3cos x = 0$
Replace in the equation (sqrt3) by $\frac{tan (π)}{3} = ⎛ ⎜ ⎝ \frac{sin (\frac{π}{3})}{\frac{cos π}{3}} ⎞ ⎟ ⎠$ $tan (pi)/3 = (sin (pi/3)/(cos pi/3))$
$sin x . cos (\frac{π}{3}) - sin (\frac{π}{3}) . cos x = 0$ $sin x.cos ((pi)/3) - sin ((pi)/3).cos x = 0$
$sin (x - \frac{π}{3}) = 0$ $sin (x - pi/3) = 0$
$(x - \frac{π}{3}) = 0 - \to$ $(x - pi/3) = 0 -->$ x = pi/3 $and$ $and$ (x - pi/3) = pi $- \to$ $-->$ x = pi + pi/3 = (4pi)/3#

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5sinx=5sqrt3cosx 0<x<2pi What are all possible solutions?

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