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UkoKoshka [18]
4 years ago
9

Solve cos (x) tan (x) -1/2=0 over the interval [0,2π].

Mathematics
1 answer:
kow [346]4 years ago
5 0
First, we are going to add \frac{1}{2} from both sides of the equation: 
cos(x)tan(x)- \frac{1}{2} + \frac{1}{2} = \frac{1}{2}
cos(x)tan(x)= \frac{1}{2}

Next, we are going to use the trig identity: tan(x)= \frac{sin(x)}{cos(x)} to rewrite our expression:
cos(x)tan(x)= \frac{1}{2}
cos(x) \frac{sin(x)}{cos(x)} = \frac{1}{2}
sin(x)= \frac{1}{2}

Finally, using our unitary circle, we can infer that sin(x)= \frac{1}{2} from 0 to 2 \pi when x=  \frac{ \pi }{6} and x= \frac{5 \pi }{6}

We can conclude that the solutions of the equation <span>cos (x) tan (x) -1/2=0 over the interval [0,2π] are: </span>x=\frac{ \pi }{6},\frac{5 \pi }{6}

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