Mathematics, 04.08.2019 02:00, zoeheskett

Ido not have a calculator to calculate this one. if you have one, . !

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Answers: x = 0, 1.05, 3.14, 5.24======sine double angle identity: $\sin 2x = 2\sin x \cos x$ apply this here $\begin{aligned} \sin 2x - \sin x & = 0 \\ 2\sin x \cos x - \sin x & = 0 \end{aligned}$ factor out $\sin x$ $\sin x(2\cos x - 1) = 0$ the left side of this equation is made up of entirely factors, namely (sin x) and (2cos x - 1). they are multiplying with each other. either factor can be equal to zero to make the entire product 0 to satisfy this equation (zero product property, like solving (x-3)(x-1) = 0)setting $\sin x$ to zero and solving: $\sin x = 0$ find an angle that has a point on the terminal arm with a y-coordinate of 0.there's two possible angles in this given interval $x=0, \pi$ note: $\pi \approx 3.14$ angle 0 radians lies directly on the positive x-axis, so y-coordinate of 0.angle pi raidans lies directly on the negative x-axis, so y-coordinate of 0.or use unit circle or the graph of y=sin x to figure this out.for the other factor: $\begin{aligned} 2\cos x - 1 & = 0 \\ 2\cos x & = 1 \\ \cos x & = \tfrac{1}{2} \end{aligned}$ we want to find the angle $x$ that has a cosine ratio of 1/2.you can use unit circle/special triangles (30-60-90) to figure the solutions out (there's two). cosine is positive in quadrants 1 and 4. with unit circle, the solutions are $\pi/3\approx 1.05$ and $5\pi/3 \approx 5.24$ alternatively, you can find the reference angle by doing, with a calculator $\begin{aligned} \cos^{-1}(\cos x) & = \cos^{-1}\left(\tfrac{1}{2}\right) \\ x_{ref} & = \cos^{-1}\left(\tfrac{1}{2}\right) \end{aligned}$ reference angle is always formed between the terminal arm and the x-axis. (refer to the attachment from my notes if you need a review on reference angles.)cosine is positive in quadrants 1 and 4. so we have a solution in quadrant 1 and quadrant 4.an angle in quadrant 1 is the reference angle. calculating using a calculator: $x_{i} = \cos^{-1}\left(\tfrac{1}{2}\right) \approx 1.05$ so that's one solutionto find an angle in quadrant 4 with a given reference angle, we need to subtract the reference angle one full rotation of $2\pi$ . reference angle is formed between the terminal arm and the x-axis; think one entire rotation and then backing up the amount of the reference angle.the angle in quadrant 4 is $x_{iv} = 2\pi - \cos^{-1}\left(\tfrac{1}{2}\right) \approx 5.24$ ======joining all the solutions: x = 0, 1.05, 3.14, 5.24
Ido not have a calculator to calculate this one. if you have one, . !