question 6\n6.1 determine the general solution for sin(x - 30°)=cos2x\n6.2 consider the functions f(x)=sin(x…

Explanation:

Step1: Use trigonometric identity

We know that $\cos2x=\sin(90^{\circ}\pm 2x)$. So, $\sin(x - 30^{\circ})=\sin(90^{\circ}\pm 2x)$. Then we have two cases: Case 1: $x-30^{\circ}=90^{\circ}+ 2x+360^{\circ}n$, where $n\in Z$. Simplify it: $x-2x=90^{\circ}+30^{\circ}+360^{\circ}n$, $-x = 120^{\circ}+360^{\circ}n$, $x=-120^{\circ}-360^{\circ}n$. Case 2: $x - 30^{\circ}=180^{\circ}-(90^{\circ}+ 2x)+360^{\circ}n$. Expand: $x - 30^{\circ}=180^{\circ}-90^{\circ}-2x+360^{\circ}n$. Combine like - terms: $x + 2x=90^{\circ}+30^{\circ}+360^{\circ}n$, $3x=120^{\circ}+360^{\circ}n$, $x = 40^{\circ}+120^{\circ}n$. So the general solution is $x=-120^{\circ}-360^{\circ}n$ or $x = 40^{\circ}+120^{\circ}n$, $n\in Z$.

Step2: Find period of $g(x)=\cos2x$

The period of the cosine function $y = A\cos(Bx + C)+D$ is $T=\frac{2\pi}{|B|}$. For $g(x)=\cos2x$, $B = 2$, so $T=\frac{2\pi}{2}=\pi$ or $180^{\circ}$.

Step3: Find range of $f(x)=\sin(x - 30^{\circ})$

The range of the sine function $y = A\sin(Bx + C)+D$ is $[D - A,D + A]$. For $f(x)=\sin(x - 30^{\circ})$, $A = 1$, $D = 0$. So the range is $[-1,1]$.

Step4: Draw graphs (description)

For $y = f(x)=\sin(x - 30^{\circ})$:

• Intercepts:
  • $x$-intercepts: $\sin(x - 30^{\circ})=0$, $x-30^{\circ}=k\cdot180^{\circ}$, $x=30^{\circ}+k\cdot180^{\circ}$. In the interval $[-90^{\circ},180^{\circ}]$, when $k=-1,x=-150^{\circ}$; when $k = 0,x = 30^{\circ}$; when $k = 1,x=210^{\circ}$ (not in the interval).
  • $y$-intercept: $x = 0$, $y=\sin(-30^{\circ})=-\frac{1}{2}$.
• Turning points: $x-30^{\circ}=(2k + 1)90^{\circ}$, $x=(2k + 1)90^{\circ}+30^{\circ}$. In the interval $[-90^{\circ},180^{\circ}]$, when $k=-1,x=-60^{\circ}$ (minimum, $y=-1$); when $k = 0,x = 120^{\circ}$ (maximum, $y = 1$). For $y = g(x)=\cos2x$:
• $x$-intercepts: $\cos2x=0$, $2x=(2k + 1)90^{\circ}$, $x=(2k + 1)45^{\circ}$. In the interval $[-90^{\circ},180^{\circ}]$, $x=-45^{\circ},45^{\circ},135^{\circ}$.
• $y$-intercept: $x = 0$, $y=\cos(0)=1$.
• Turning points: $2x=k\cdot180^{\circ}$, $x=k\cdot90^{\circ}$. In the interval $[-90^{\circ},180^{\circ}]$, when $k=-1,x=-90^{\circ}$ (minimum, $y=-1$); when $k = 0,x = 0$ (maximum, $y = 1$); when $k = 1,x=90^{\circ}$ (minimum, $y=-1$); when $k = 2,x=180^{\circ}$ (maximum, $y = 1$).

Step5: Find intersection points

We already found the general solution in Step1. In the interval $x\in[-90^{\circ},180^{\circ}]$: For $x=-120^{\circ}-360^{\circ}n$, when $n = 0,x=-120^{\circ}$ (not in the interval). For $x = 40^{\circ}+120^{\circ}n$, when $n=-1,x=-80^{\circ}$; when $n = 0,x = 40^{\circ}$; when $n = 1,x=160^{\circ}$.

Answer:

6.1: $x=-120^{\circ}-360^{\circ}n$ or $x = 40^{\circ}+120^{\circ}n$, $n\in Z$ 6.2.1: $180^{\circ}$ 6.2.2: $[-1,1]$ 6.2.3: (Graph should be drawn as described above) 6.2.4: $-80^{\circ},40^{\circ},160^{\circ}$