Question #40645

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Question #40645

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Answer 1 · 2016-05-09T23:23:07

$x = \frac{7 π}{6} + 2 k π, \frac{11 π}{6} + 2 k π$ $x=(7pi)/6+2kpi,(11pi)/6+2kpi$ , where $k$ $k$ is an integer

Explanation:

We have the equation:

$15 sin x + 7 = sin x$ $15sinx+7=sinx$

Subtract $7$ $7$ from both sides.

$15 sin x = sin x - 7$ $15sinx=sinx-7$

Subtract $sin x$ $sinx$ from both sides.

$14 sin x = - 7$ $14sinx=-7$

Divide both sides by $14$ $14$ .

$sin x = - \frac{1}{2}$ $sinx=-1/2$

If we are restricting our domain from $0 \leq x < 2 π$ $0<=x<2pi$ , our solutions are at the reference angles of $\frac{π}{6}$ $pi/6$ in $QIII$ $"QIII"$ and $QIV$ $"QIV"$ , when $sin x$ $sinx$ is negative. These values give us

$x = \frac{7 π}{6}, \frac{11 π}{6}$ $x=(7pi)/6,(11pi)/6$

However, if we don't restrict our domain, these two values and any angle $2 π$ $2pi$ away will also satisfy the equation, thus the full solution is:

$x = \frac{7 π}{6} + 2 k π, \frac{11 π}{6} + 2 k π$ $x=(7pi)/6+2kpi,(11pi)/6+2kpi$ , where $k$ $k$ is an integer

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Question #40645

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