Math  /  Trigonometry

Question8. Using a calculator, determine the solutions for each equation, to two decimal places, on the interval 0x2π0 \leq x \leq 2 \pi. a) 3sinx=sinx+13 \sin x=\sin x+1 c) cosx1=cosx\cos x-1=-\cos x b) 5cosx3=3cosx5 \cos x-\sqrt{3}=3 \cos x d) 5sinx+1=3sinx5 \sin x+1=3 \sin x

Studdy Solution
Finally, solve the equation 5sinx+1=3sinx5 \sin x + 1 = 3 \sin x.
Subtract 3sinx3 \sin x from both sides:
5sinx3sinx=1 5 \sin x - 3 \sin x = -1 2sinx=1 2 \sin x = -1
Divide both sides by 2:
sinx=12 \sin x = -\frac{1}{2}
Using a calculator, find xx such that sinx=12\sin x = -\frac{1}{2} in the interval 0x2π0 \leq x \leq 2\pi.
The solutions are:
x=7π6,11π6 x = \frac{7\pi}{6}, \frac{11\pi}{6}
The solutions for each equation are: a) x=π6,5π6 x = \frac{\pi}{6}, \frac{5\pi}{6} b) x=π3,5π3 x = \frac{\pi}{3}, \frac{5\pi}{3} c) x=π6,11π6 x = \frac{\pi}{6}, \frac{11\pi}{6} d) x=7π6,11π6 x = \frac{7\pi}{6}, \frac{11\pi}{6}

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