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In two places I'd written "are integer multiples of $2n\pi$ added to the above", which although is technically correct, is unnecessarily convoluted.
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Dave L Renfro
Dave L Renfro
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This was written after the OP's Edit update, which in turn was made after Xander Henderson's answer.

Using $\;a \cos x + b \sin x \; = \;R \sin (x + \alpha)$

Assuming the student is allowed to use without verification the relevant formulas, maybe this would be how I'd want students to present their work. (Note: All decimal expansions are 3-digit truncations of the corresponding exact values.)

Since $\;6\cos x - 8\sin x = R\sin(x + \alpha)\;$ for $R = -\sqrt{6^2 + 8^2} = -10$ and $\alpha = \arctan \left(-\frac{3}{4}\right) = -0.643 \ldots,$ we have

$$-10 \sin \left[x \; + \; \arctan\left(-\frac{3}{4}\right)\right] \;\; = \;\; 7$$

$$\sin \left[x \; - \; \arctan\left(\frac{3}{4}\right)\right] \;\; = \;\; -0.7$$

Sine is negative only in Q3 and Q4, and because $\arcsin(-0.7) = -0.775 \ldots,$ the $(x$-axis) reference angle for $x - \arctan\left(\frac{3}{4}\right)$ in these quadrants is $0.775 \ldots$. Therefore,

$$ x \; - \; \arctan\left(\frac{3}{4}\right) \; = \; \pi + 0.775\ldots, \;\; 2\pi - 0.775\ldots $$

$$ x \; - \; \arctan\left(\frac{3}{4}\right) \; = \; 3.916 \ldots, \;\; 5.507 \ldots $$

$$ x \;\; = \;\; (3.916 \ldots) + (0.643 \ldots), \;\; (5.507 \ldots) + (0.643 \ldots) $$

$$ x \;\; = \;\; 4.560 \ldots, \;\; 6.151 \ldots $$

Since all solutions to $\;\sin (\text{stuff}) = -0.7\;$ are integer multiples ofgiven by adding $2n\pi$ added to the above two solutions, (i.e. by adding integer multiples of $2\pi),$ and only $n=0$ gives values of $x$ such that $0 < x < 2\pi,$ the above two values are all solutions for $x$ such that $0 < x < 2\pi.$

Using $\;\sin^2 x + \cos^2 x = 1$

$$ 6\cos x \; = \; 7 + 8\sin x $$

$$ (6\cos x)^2 \; = \; (7 + 8\sin x)^2 $$

Because we've squared both sides of an equation, we will need to check for extraneous solutions at the end.

$$ 36\cos^2 x \;\; = \;\; 49 + 112\sin x + 64\sin^2 x $$

$$ 36(1 - \sin^2 x) \;\; = \;\; 49 + 112\sin x + 64\sin^2 x $$

$$ 100\sin^2 x + 112\sin x + 13 \; = \; 0 $$

$$ \sin x \;\; = \;\; \frac{-112 \; \pm \; \sqrt{{112}^2 \; - \; 4(100)(13)}}{200} $$

$$ \sin x \;\; = \;\; \frac{-112 \; \pm \; \sqrt{7344}}{200} $$

$$ \sin x \;\; = \;\; -0.131 \ldots, \;\; -0.988 \ldots $$

Sine is negative only in Q3 and Q4, and because $\arcsin(-0.131 \ldots) = -0.131 \ldots$ and $\arcsin(-0.988 \ldots) = -1.418 \ldots,$ the $(x$-axis) reference angles for $x$ in these quadrants are $0.131 \ldots$ and $1.418 \ldots$. Therefore,

$$ x \;\; = \;\; \pi + 0.131 \ldots, \;\; \pi + 1.418 \ldots, \;\; 2\pi - 0.131 \ldots, \;\; 2\pi - 1.418 \ldots $$

$$ x \;\; = \;\; 3.273 \ldots, \;\; 4.560 \ldots, \;\; 6.151 \ldots, \;\; 4.864 \ldots $$

Since all solutions to $\;\sin (\text{stuff}) = \text{constant}\;$ are integer multiples ofgiven by adding $2n\pi$ added to the above four solutions, (i.e. by adding integer multiples of $2\pi),$ and only $n=0$ gives values of $x$ such that $0 < x < 2\pi,$ the above four values include all solutions for $x$ such that $0 < x < 2\pi.$

Checking for extraneous solutions, we find that $3.273 \ldots$ and $4.864 \ldots$ do not satisfy the original equation. Therefore, the solutions to the given equation such that $0 < x < 2\pi$ are $ x = 4.560 \ldots$ and $x = 6.151 \ldots$.

This was written after the OP's Edit update, which in turn was made after Xander Henderson's answer.

Using $\;a \cos x + b \sin x \; = \;R \sin (x + \alpha)$

Assuming the student is allowed to use without verification the relevant formulas, maybe this would be how I'd want students to present their work. (Note: All decimal expansions are 3-digit truncations of the corresponding exact values.)

Since $\;6\cos x - 8\sin x = R\sin(x + \alpha)\;$ for $R = -\sqrt{6^2 + 8^2} = -10$ and $\alpha = \arctan \left(-\frac{3}{4}\right) = -0.643 \ldots,$ we have

$$-10 \sin \left[x \; + \; \arctan\left(-\frac{3}{4}\right)\right] \;\; = \;\; 7$$

$$\sin \left[x \; - \; \arctan\left(\frac{3}{4}\right)\right] \;\; = \;\; -0.7$$

Sine is negative only in Q3 and Q4, and because $\arcsin(-0.7) = -0.775 \ldots,$ the $(x$-axis) reference angle for $x - \arctan\left(\frac{3}{4}\right)$ in these quadrants is $0.775 \ldots$. Therefore,

$$ x \; - \; \arctan\left(\frac{3}{4}\right) \; = \; \pi + 0.775\ldots, \;\; 2\pi - 0.775\ldots $$

$$ x \; - \; \arctan\left(\frac{3}{4}\right) \; = \; 3.916 \ldots, \;\; 5.507 \ldots $$

$$ x \;\; = \;\; (3.916 \ldots) + (0.643 \ldots), \;\; (5.507 \ldots) + (0.643 \ldots) $$

$$ x \;\; = \;\; 4.560 \ldots, \;\; 6.151 \ldots $$

Since all solutions to $\;\sin (\text{stuff}) = -0.7\;$ are integer multiples of $2n\pi$ added to the above two solutions, and only $n=0$ gives values of $x$ such that $0 < x < 2\pi,$ the above two values are all solutions for $x$ such that $0 < x < 2\pi.$

Using $\;\sin^2 x + \cos^2 x = 1$

$$ 6\cos x \; = \; 7 + 8\sin x $$

$$ (6\cos x)^2 \; = \; (7 + 8\sin x)^2 $$

Because we've squared both sides of an equation, we will need to check for extraneous solutions at the end.

$$ 36\cos^2 x \;\; = \;\; 49 + 112\sin x + 64\sin^2 x $$

$$ 36(1 - \sin^2 x) \;\; = \;\; 49 + 112\sin x + 64\sin^2 x $$

$$ 100\sin^2 x + 112\sin x + 13 \; = \; 0 $$

$$ \sin x \;\; = \;\; \frac{-112 \; \pm \; \sqrt{{112}^2 \; - \; 4(100)(13)}}{200} $$

$$ \sin x \;\; = \;\; \frac{-112 \; \pm \; \sqrt{7344}}{200} $$

$$ \sin x \;\; = \;\; -0.131 \ldots, \;\; -0.988 \ldots $$

Sine is negative only in Q3 and Q4, and because $\arcsin(-0.131 \ldots) = -0.131 \ldots$ and $\arcsin(-0.988 \ldots) = -1.418 \ldots,$ the $(x$-axis) reference angles for $x$ in these quadrants are $0.131 \ldots$ and $1.418 \ldots$. Therefore,

$$ x \;\; = \;\; \pi + 0.131 \ldots, \;\; \pi + 1.418 \ldots, \;\; 2\pi - 0.131 \ldots, \;\; 2\pi - 1.418 \ldots $$

$$ x \;\; = \;\; 3.273 \ldots, \;\; 4.560 \ldots, \;\; 6.151 \ldots, \;\; 4.864 \ldots $$

Since all solutions to $\;\sin (\text{stuff}) = \text{constant}\;$ are integer multiples of $2n\pi$ added to the above four solutions, and only $n=0$ gives values of $x$ such that $0 < x < 2\pi,$ the above four values include all solutions for $x$ such that $0 < x < 2\pi.$

Checking for extraneous solutions, we find that $3.273 \ldots$ and $4.864 \ldots$ do not satisfy the original equation. Therefore, the solutions to the given equation such that $0 < x < 2\pi$ are $ x = 4.560 \ldots$ and $x = 6.151 \ldots$.

This was written after the OP's Edit update, which in turn was made after Xander Henderson's answer.

Using $\;a \cos x + b \sin x \; = \;R \sin (x + \alpha)$

Assuming the student is allowed to use without verification the relevant formulas, maybe this would be how I'd want students to present their work. (Note: All decimal expansions are 3-digit truncations of the corresponding exact values.)

Since $\;6\cos x - 8\sin x = R\sin(x + \alpha)\;$ for $R = -\sqrt{6^2 + 8^2} = -10$ and $\alpha = \arctan \left(-\frac{3}{4}\right) = -0.643 \ldots,$ we have

$$-10 \sin \left[x \; + \; \arctan\left(-\frac{3}{4}\right)\right] \;\; = \;\; 7$$

$$\sin \left[x \; - \; \arctan\left(\frac{3}{4}\right)\right] \;\; = \;\; -0.7$$

Sine is negative only in Q3 and Q4, and because $\arcsin(-0.7) = -0.775 \ldots,$ the $(x$-axis) reference angle for $x - \arctan\left(\frac{3}{4}\right)$ in these quadrants is $0.775 \ldots$. Therefore,

$$ x \; - \; \arctan\left(\frac{3}{4}\right) \; = \; \pi + 0.775\ldots, \;\; 2\pi - 0.775\ldots $$

$$ x \; - \; \arctan\left(\frac{3}{4}\right) \; = \; 3.916 \ldots, \;\; 5.507 \ldots $$

$$ x \;\; = \;\; (3.916 \ldots) + (0.643 \ldots), \;\; (5.507 \ldots) + (0.643 \ldots) $$

$$ x \;\; = \;\; 4.560 \ldots, \;\; 6.151 \ldots $$

Since all solutions to $\;\sin (\text{stuff}) = -0.7\;$ are given by adding $2n\pi$ to the above two solutions (i.e. by adding integer multiples of $2\pi),$ and only $n=0$ gives values of $x$ such that $0 < x < 2\pi,$ the above two values are all solutions for $x$ such that $0 < x < 2\pi.$

Using $\;\sin^2 x + \cos^2 x = 1$

$$ 6\cos x \; = \; 7 + 8\sin x $$

$$ (6\cos x)^2 \; = \; (7 + 8\sin x)^2 $$

Because we've squared both sides of an equation, we will need to check for extraneous solutions at the end.

$$ 36\cos^2 x \;\; = \;\; 49 + 112\sin x + 64\sin^2 x $$

$$ 36(1 - \sin^2 x) \;\; = \;\; 49 + 112\sin x + 64\sin^2 x $$

$$ 100\sin^2 x + 112\sin x + 13 \; = \; 0 $$

$$ \sin x \;\; = \;\; \frac{-112 \; \pm \; \sqrt{{112}^2 \; - \; 4(100)(13)}}{200} $$

$$ \sin x \;\; = \;\; \frac{-112 \; \pm \; \sqrt{7344}}{200} $$

$$ \sin x \;\; = \;\; -0.131 \ldots, \;\; -0.988 \ldots $$

Sine is negative only in Q3 and Q4, and because $\arcsin(-0.131 \ldots) = -0.131 \ldots$ and $\arcsin(-0.988 \ldots) = -1.418 \ldots,$ the $(x$-axis) reference angles for $x$ in these quadrants are $0.131 \ldots$ and $1.418 \ldots$. Therefore,

$$ x \;\; = \;\; \pi + 0.131 \ldots, \;\; \pi + 1.418 \ldots, \;\; 2\pi - 0.131 \ldots, \;\; 2\pi - 1.418 \ldots $$

$$ x \;\; = \;\; 3.273 \ldots, \;\; 4.560 \ldots, \;\; 6.151 \ldots, \;\; 4.864 \ldots $$

Since all solutions to $\;\sin (\text{stuff}) = \text{constant}\;$ are given by adding $2n\pi$ to the above four solutions (i.e. by adding integer multiples of $2\pi),$ and only $n=0$ gives values of $x$ such that $0 < x < 2\pi,$ the above four values include all solutions for $x$ such that $0 < x < 2\pi.$

Checking for extraneous solutions, we find that $3.273 \ldots$ and $4.864 \ldots$ do not satisfy the original equation. Therefore, the solutions to the given equation such that $0 < x < 2\pi$ are $ x = 4.560 \ldots$ and $x = 6.151 \ldots$.

Source Link
Dave L Renfro
Dave L Renfro
  • 6.1k
  • 1
  • 17
  • 29

This was written after the OP's Edit update, which in turn was made after Xander Henderson's answer.

Using $\;a \cos x + b \sin x \; = \;R \sin (x + \alpha)$

Assuming the student is allowed to use without verification the relevant formulas, maybe this would be how I'd want students to present their work. (Note: All decimal expansions are 3-digit truncations of the corresponding exact values.)

Since $\;6\cos x - 8\sin x = R\sin(x + \alpha)\;$ for $R = -\sqrt{6^2 + 8^2} = -10$ and $\alpha = \arctan \left(-\frac{3}{4}\right) = -0.643 \ldots,$ we have

$$-10 \sin \left[x \; + \; \arctan\left(-\frac{3}{4}\right)\right] \;\; = \;\; 7$$

$$\sin \left[x \; - \; \arctan\left(\frac{3}{4}\right)\right] \;\; = \;\; -0.7$$

Sine is negative only in Q3 and Q4, and because $\arcsin(-0.7) = -0.775 \ldots,$ the $(x$-axis) reference angle for $x - \arctan\left(\frac{3}{4}\right)$ in these quadrants is $0.775 \ldots$. Therefore,

$$ x \; - \; \arctan\left(\frac{3}{4}\right) \; = \; \pi + 0.775\ldots, \;\; 2\pi - 0.775\ldots $$

$$ x \; - \; \arctan\left(\frac{3}{4}\right) \; = \; 3.916 \ldots, \;\; 5.507 \ldots $$

$$ x \;\; = \;\; (3.916 \ldots) + (0.643 \ldots), \;\; (5.507 \ldots) + (0.643 \ldots) $$

$$ x \;\; = \;\; 4.560 \ldots, \;\; 6.151 \ldots $$

Since all solutions to $\;\sin (\text{stuff}) = -0.7\;$ are integer multiples of $2n\pi$ added to the above two solutions, and only $n=0$ gives values of $x$ such that $0 < x < 2\pi,$ the above two values are all solutions for $x$ such that $0 < x < 2\pi.$

Using $\;\sin^2 x + \cos^2 x = 1$

$$ 6\cos x \; = \; 7 + 8\sin x $$

$$ (6\cos x)^2 \; = \; (7 + 8\sin x)^2 $$

Because we've squared both sides of an equation, we will need to check for extraneous solutions at the end.

$$ 36\cos^2 x \;\; = \;\; 49 + 112\sin x + 64\sin^2 x $$

$$ 36(1 - \sin^2 x) \;\; = \;\; 49 + 112\sin x + 64\sin^2 x $$

$$ 100\sin^2 x + 112\sin x + 13 \; = \; 0 $$

$$ \sin x \;\; = \;\; \frac{-112 \; \pm \; \sqrt{{112}^2 \; - \; 4(100)(13)}}{200} $$

$$ \sin x \;\; = \;\; \frac{-112 \; \pm \; \sqrt{7344}}{200} $$

$$ \sin x \;\; = \;\; -0.131 \ldots, \;\; -0.988 \ldots $$

Sine is negative only in Q3 and Q4, and because $\arcsin(-0.131 \ldots) = -0.131 \ldots$ and $\arcsin(-0.988 \ldots) = -1.418 \ldots,$ the $(x$-axis) reference angles for $x$ in these quadrants are $0.131 \ldots$ and $1.418 \ldots$. Therefore,

$$ x \;\; = \;\; \pi + 0.131 \ldots, \;\; \pi + 1.418 \ldots, \;\; 2\pi - 0.131 \ldots, \;\; 2\pi - 1.418 \ldots $$

$$ x \;\; = \;\; 3.273 \ldots, \;\; 4.560 \ldots, \;\; 6.151 \ldots, \;\; 4.864 \ldots $$

Since all solutions to $\;\sin (\text{stuff}) = \text{constant}\;$ are integer multiples of $2n\pi$ added to the above four solutions, and only $n=0$ gives values of $x$ such that $0 < x < 2\pi,$ the above four values include all solutions for $x$ such that $0 < x < 2\pi.$

Checking for extraneous solutions, we find that $3.273 \ldots$ and $4.864 \ldots$ do not satisfy the original equation. Therefore, the solutions to the given equation such that $0 < x < 2\pi$ are $ x = 4.560 \ldots$ and $x = 6.151 \ldots$.