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Math  /  Trigonometry

Question3) Solve the following trigonometric equations within the domain [0,2π][0,2 \pi]. Round to the nearest tenth and use radians.  a) 2sin⁡2x−1=02sin⁡3x2=12sin⁡x=±12 or \text { a) } \begin{array}{l} 2 \sin ^{2} x-1=0 \\ \frac{2 \sin ^{3} x}{2}=\frac{1}{2} \\ \sin x= \pm \frac{1}{\sqrt{2}} \\ \text { or } \end{array} (2sin⁡x−1)(2sin⁡x+1)(\sqrt{2} \sin x-1)(\sqrt{2} \sin x+1) 2sin⁡x=12sin⁡x=−1\sqrt{2} \sin x=1 \quad \sqrt{2} \sin x=-1 sin⁡x=12sin⁡x=−12\sin x=\frac{1}{\sqrt{2}} \quad \sin x=-\frac{1}{\sqrt{2}} x=π4x=π4x=\frac{\pi}{4} \quad x=\frac{\pi}{4}  b) csc⁡2x+2csc⁡x−3=0csc⁡2x−1csc⁡x+3csc⁡x−3=0csc⁡x(csc⁡x−1)3(csc⁡x−1)=0csc⁡x+3=0csc⁡x−1=0csc⁡x=−3csc⁡x=11sin⁡x=−31sin⁡x=1sin⁡x=1−3sin⁡x=11sin⁡−1x=0.340x=π2\begin{array}{ll} \text { b) } \csc ^{2} x+2 \csc x-3=0 \\ \csc ^{2} x-1 \csc x+3 \csc x-3=0 \\ \csc x(\csc x-1) & 3(\csc x-1)=0 \\ \csc x+3=0 & \csc x-1=0 \\ \csc x=-3 & \csc x=1 \\ \frac{1}{\sin _{x}}=-3 & \frac{1}{\sin x}=1 \\ \sin x=\frac{1}{-3} & \sin x=\frac{1}{1} \\ \sin ^{-1} x=0.340 & x=\frac{\pi}{2} \end{array} c) cos⁡x−cos⁡2x=1/2\cos x-\cos 2 x=1 / 2

Studdy Solution

STEP 1

1. We are solving trigonometric equations within the domain [0,2π][0, 2\pi].
2. Solutions should be rounded to the nearest tenth and expressed in radians.

STEP 2

1. Solve equation (a) for sin⁡x \sin x .
2. Solve equation (b) for csc⁡x \csc x .
3. Solve equation (c) for cos⁡x \cos x .

STEP 3

For equation (a), start with the given equation:
2sin⁡2x−1=0 2 \sin^2 x - 1 = 0
Rearrange the equation to solve for sin⁡2x\sin^2 x:
2sin⁡2x=1 2 \sin^2 x = 1 sin⁡2x=12 \sin^2 x = \frac{1}{2}
Take the square root of both sides to solve for sin⁡x\sin x:
sin⁡x=±12 \sin x = \pm \frac{1}{\sqrt{2}}

STEP 4

Find the values of x x within the domain [0,2π][0, 2\pi] where sin⁡x=12\sin x = \frac{1}{\sqrt{2}} and sin⁡x=−12\sin x = -\frac{1}{\sqrt{2}}.
For sin⁡x=12\sin x = \frac{1}{\sqrt{2}}:
x=π4,3π4 x = \frac{\pi}{4}, \frac{3\pi}{4}
For sin⁡x=−12\sin x = -\frac{1}{\sqrt{2}}:
x=5π4,7π4 x = \frac{5\pi}{4}, \frac{7\pi}{4}

STEP 5

For equation (b), start with the given equation:
csc⁡2x+2csc⁡x−3=0 \csc^2 x + 2 \csc x - 3 = 0
Let y=csc⁡x y = \csc x . The equation becomes:
y2+2y−3=0 y^2 + 2y - 3 = 0
Factor the quadratic equation:
(y+3)(y−1)=0 (y + 3)(y - 1) = 0
Solve for y y :
y=−3ory=1 y = -3 \quad \text{or} \quad y = 1
Since y=csc⁡x y = \csc x , we have:
csc⁡x=−3orcsc⁡x=1 \csc x = -3 \quad \text{or} \quad \csc x = 1

STEP 6

Convert csc⁡x\csc x to sin⁡x\sin x:
For csc⁡x=−3\csc x = -3:
sin⁡x=−13 \sin x = -\frac{1}{3}
For csc⁡x=1\csc x = 1:
sin⁡x=1 \sin x = 1
Find the values of x x within the domain [0,2π][0, 2\pi]:
For sin⁡x=1\sin x = 1:
x=π2 x = \frac{\pi}{2}
For sin⁡x=−13\sin x = -\frac{1}{3}, use a calculator to find:
x≈3.48 radians x \approx 3.48 \, \text{radians}

STEP 7

For equation (c), start with the given equation:
cos⁡x−cos⁡2x=12 \cos x - \cos 2x = \frac{1}{2}
Use the double angle identity cos⁡2x=2cos⁡2x−1\cos 2x = 2\cos^2 x - 1 to rewrite the equation:
cos⁡x−(2cos⁡2x−1)=12 \cos x - (2\cos^2 x - 1) = \frac{1}{2}
Simplify:
cos⁡x−2cos⁡2x+1=12 \cos x - 2\cos^2 x + 1 = \frac{1}{2}
Rearrange the equation:
−2cos⁡2x+cos⁡x+12=0 -2\cos^2 x + \cos x + \frac{1}{2} = 0
Multiply through by −1-1 to make it easier to factor:
2cos⁡2x−cos⁡x−12=0 2\cos^2 x - \cos x - \frac{1}{2} = 0

STEP 8

Factor the quadratic equation:
(2cos⁡x+1)(cos⁡x−12)=0 (2\cos x + 1)(\cos x - \frac{1}{2}) = 0
Solve for cos⁡x\cos x:
cos⁡x=−12orcos⁡x=12 \cos x = -\frac{1}{2} \quad \text{or} \quad \cos x = \frac{1}{2}
Find the values of x x within the domain [0,2π][0, 2\pi]:
For cos⁡x=−12\cos x = -\frac{1}{2}:
x=2π3,4π3 x = \frac{2\pi}{3}, \frac{4\pi}{3}
For cos⁡x=12\cos x = \frac{1}{2}:
x=π3,5π3 x = \frac{\pi}{3}, \frac{5\pi}{3}
The solutions for each part are:
a) x=π4,3π4,5π4,7π4 x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}
b) x≈3.48,π2 x \approx 3.48, \frac{\pi}{2}
c) x=2π3,4π3,π3,5π3 x = \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{\pi}{3}, \frac{5\pi}{3}

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