Math  /  Trigonometry

Question12. Use an algebraic method to solve the following equations. Show all your work for full marks. a. 2sin(2x)1=02 \sin (2 x)-1=0, where x[0,2π]x \in[0,2 \pi]. (4)

Studdy Solution

STEP 1

1. We are solving the trigonometric equation 2sin(2x)1=0 2 \sin(2x) - 1 = 0 .
2. The solution should be within the interval x[0,2π] x \in [0, 2\pi] .
3. We will use algebraic manipulation and trigonometric identities to solve the equation.

STEP 2

1. Isolate the trigonometric function.
2. Solve for the angle 2x 2x .
3. Find the solutions for x x within the given interval.

STEP 3

First, isolate the sine function by adding 1 to both sides and then dividing by 2:
2sin(2x)1=0 2 \sin(2x) - 1 = 0 2sin(2x)=1 2 \sin(2x) = 1 sin(2x)=12 \sin(2x) = \frac{1}{2}

STEP 4

Next, solve for 2x 2x by finding the angles whose sine is 12\frac{1}{2}. The sine function equals 12\frac{1}{2} at angles π6\frac{\pi}{6} and 5π6\frac{5\pi}{6} within one period [0,2π][0, 2\pi]. Therefore, the general solutions for 2x 2x are:
2x=π6+2kπand2x=5π6+2kπ 2x = \frac{\pi}{6} + 2k\pi \quad \text{and} \quad 2x = \frac{5\pi}{6} + 2k\pi
where k k is any integer.

STEP 5

Since the sine function is periodic with period 2π 2\pi , we also consider the solutions within one period:
2x=π6+2kπx=π12+kπ 2x = \frac{\pi}{6} + 2k\pi \quad \Rightarrow \quad x = \frac{\pi}{12} + k\pi 2x=5π6+2kπx=5π12+kπ 2x = \frac{5\pi}{6} + 2k\pi \quad \Rightarrow \quad x = \frac{5\pi}{12} + k\pi

STEP 6

Determine the values of x x within the interval [0,2π][0, 2\pi]. For each case, calculate the possible values of x x by substituting integer values for k k :
1. For x=π12+kπ x = \frac{\pi}{12} + k\pi : - k=0 k = 0 : x=π12 x = \frac{\pi}{12} - k=1 k = 1 : x=π12+π=13π12 x = \frac{\pi}{12} + \pi = \frac{13\pi}{12}
2. For x=5π12+kπ x = \frac{5\pi}{12} + k\pi : - k=0 k = 0 : x=5π12 x = \frac{5\pi}{12} - k=1 k = 1 : x=5π12+π=17π12 x = \frac{5\pi}{12} + \pi = \frac{17\pi}{12}
Thus, the solutions for x x within the interval [0,2π][0, 2\pi] are:
x=π12,13π12,5π12,17π12 x = \frac{\pi}{12}, \frac{13\pi}{12}, \frac{5\pi}{12}, \frac{17\pi}{12}
The solutions for x x are:
π12,13π12,5π12,17π12 \boxed{\frac{\pi}{12}, \frac{13\pi}{12}, \frac{5\pi}{12}, \frac{17\pi}{12}}

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