Find an angle $x$ that satisfies the equation: $$3\cdot\cos\left(\frac{x}{3}+30^\circ\right)=4-7\cdot\cos\left(\frac{x}{3}+30^\circ\right)$$ Use a calculator for angle calculations. Round the resulting angle to one decimal place.
Jana's solution:
(1) By moving all cosine terms to one side of the equation and combining them, Jana obtained the simplified equation: \begin{aligned} 10\cdot\cos\left(\frac{x}{3}+30^\circ\right)&=4\cr \cos\left(\frac{x}{3}+30^\circ\right)&=0.4 \end{aligned} (2) Using the substitution $b=\frac{x}{3}+30^\circ$ she further simplified the equation to: $$\cos b=0.4$$ (3) She then used a calculator to solve this basic trigonometric equation. After rounding to one decimal place, she determined the value of the substituted variable as: $$b\approx 66.4^\circ+k\cdot360^\circ,\mbox{ where } k\in\mathbb{Z}$$ (4) Finally, after substituting back and solving for $x$, she found: \begin{aligned} \frac{x}{3}+30^\circ&=66.4^\circ+k\cdot360^\circ\cr x&\approx109.2^\circ+k\cdot 1\,080^\circ,\mbox{ where } k\in\mathbb{Z} \end{aligned} However, Jana made a mistake in one step. Identify the incorrect step.
The mistake is in step (1). Jana made a mistake when simplifying the equation. The correct simplified equation should have been: $$-4\cdot\cos\left(\frac{x}{3}+30^\circ\right)=4$$
The mistake is in step (2). The substitution should have been: $$b=\cos\left(\frac{x}{3}+30^\circ\right)$$ Then, the substituted variable would be $b=0.4$.
The mistake is in step (3). The equation $\cos b=0.4$ has two solutions on the interval $[ 0^\circ;360^\circ ]$.
The mistake is in step (4). When solving for $x$, Jana incorrectly determined the period. The correct result should have been. \begin{aligned} \frac{x}{3}+30^\circ&\approx66.4^\circ+k\cdot360^\circ\cr x&\approx 109.2^\circ+k\cdot360^\circ,\mbox{ where }k\in\mathbb{Z} \end{aligned}
The equation $\cos b=0.4$ has two solutions on the interval $[ 0^\circ;360^\circ ]$. However, the calculator only displays the solution from the first quadrant!
In addition to the first-quadrant solution: $$b_1\approx 66.4^\circ+k\cdot360^\circ$$ we must also include the fourth-quadrant solution: $$b_2\approx\left(360^\circ-66.4^\circ\right)+k\cdot360^\circ$$ Finally, we must return to the substitution and solve for the unknown $x$:
For the first-quadrant solution: \begin{aligned} b_1&\approx 66.4^\circ+k\cdot360^\circ,\mbox{ where } k\in\mathbb{Z}\cr \frac{x_1}{3}+30^\circ&\approx 66.4^\circ+k\cdot 360^\circ\cr \frac{x_1}{3}&\approx 36.4^\circ+k\cdot360^\circ\cr x_1&\approx 109.2^\circ+k\cdot 1\,080^\circ \end{aligned} For the fourth-quadrant solution: \begin{aligned} b_2&\approx293.6^\circ+k\cdot360^\circ,\mbox{ where } k\in\mathbb{Z}\cr \frac{x_2}{3}+30^\circ&\approx 293.6^\circ+k\cdot360^\circ\cr \frac{x_2}{3}&\approx263.6^\circ+k\cdot360^\circ\cr x_2&\approx790.8^\circ+k\cdot 1\,080^\circ \end{aligned} The correct result in proper form is: \begin{aligned} x_1&\approx 109.2^\circ+k\cdot 1\,080^\circ,\mbox{ where }k\in\mathbb{Z}\cr x_2&\approx 790.8^\circ+k\cdot 1\,080^\circ,\mbox{ where }k\in\mathbb{Z} \end{aligned}