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) 2sin2x1=02sin3x2=12sinx=±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} (2sinx1)(2sinx+1)(\sqrt{2} \sin x-1)(\sqrt{2} \sin x+1) 2sinx=12sinx=1\sqrt{2} \sin x=1 \quad \sqrt{2} \sin x=-1 sinx=12sinx=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) csc2x+2cscx3=0csc2x1cscx+3cscx3=0cscx(cscx1)3(cscx1)=0cscx+3=0cscx1=0cscx=3cscx=11sinx=31sinx=1sinx=13sinx=11sin1x=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) cosxcos2x=1/2\cos x-\cos 2 x=1 / 2

Studdy Solution
Factor the quadratic equation:
(2cosx+1)(cosx12)=0 (2\cos x + 1)(\cos x - \frac{1}{2}) = 0
Solve for cosx\cos x:
cosx=12orcosx=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 cosx=12\cos x = -\frac{1}{2}:
x=2π3,4π3 x = \frac{2\pi}{3}, \frac{4\pi}{3}
For cosx=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) x3.48,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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