How do you solve the equation $cosZ=0.98$ ?

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Answer 1 · 2018-01-12T00:20:37

$Z = cos^(-1)(0.98) + 2pi*n$ , $n in ZZ$ .

or

$Z=- cos^(-1)(0.98) + 2pi*n$ , $n in ZZ$ .

Since $Z>0$ we know that $Z$ could come from QI or QIV. We'll find $hatZ$ , the reference angle for $Z$ :

$cos(Z)=0.98 rarr hatZ=cos^(-1)(0.98)$

For the QI angle, we have $Z = cos^(-1)(0.98) + 2pi*n$ , $n in ZZ$ .

For QIV, we have $Z = 2pi - cos^(-1)(0.98) + 2pi*n$ , $n in ZZ$ , which could be rewritten: $- cos^(-1)(0.98) + 2pi*n$ , $n in ZZ$ .

How do you solve the equation cosZ=0.98cosZ=0.98?