$sin (60 + x) - sin (60 - x) = 1$ $sin(60+x)-sin(60-x)=1$ the interval $0 < x < 2 π$ $0<x<2pi$ how do you solve?

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Answer 1 · 2018-02-17T14:54:14

$x = \frac{π}{2}$ $x=pi/2$

We want to solve the equation

$sin (60^{\circ} + x) - sin (60^{\circ} - x) = 1$ $sin(60^@+x)-sin(60^@-x)=1$

We will use the identity

$sin (a + b) - sin (a - b) = 2 sin (b) cos (a)$ $sin(a+b)-sin(a-b)=2sin(b)cos(a)$

(Can be derived using angle-sum and difference identities)

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Apply the identity ( $a = 60$ $a=60$ and $b = x$ $b=x$ )

$2 sin (x) cos (60^{\circ}) = 1$ $2sin(x)cos(60^@)=1$

Solve for sine

$sin (x) = \frac{1}{2 cos (60^{\circ})}$ $sin(x)=1/(2cos(60^@))$

$= \frac{1}{2 \cdot \frac{1}{2}}$ $=1/(2*1/2)$

$= 1$ $=1$

By the inverse sine

$x = \frac{π}{2}$ $x=pi/2$

sin(60+x)−sin(60−x)=1sin(60+x)-sin(60-x)=1 the interval 0<x<2π0<x<2pi how do you solve?