Solving Trigonomic Equation. IF 0<x<360, what is the value of cosx in the equation 2cos^2x+sin^2x=41/25?

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Answer 1 · 2018-01-06T18:20:53

$cos (x) = \pm \frac{4}{5}$ $cos(x) = pm 4/5$

We know that ${cos}^{2} (x) + {sin}^{2} (x) = 1 \to {sin}^{2} (x) = 1 - {cos}^{2} (x)$ $cos^2(x)+sin^2(x)=1\rightarrow sin^2(x) = 1-cos^2(x)$ .

Substituting we have:

$2 {cos}^{2} (x) + (1 - {cos}^{2} (x)) = \frac{41}{25}$ $2cos^2(x)+(1-cos^2(x))=41/25$

${cos}^{2} (x) + 1 = \frac{41}{25}$ $cos^2(x)+1=41/25$

${cos}^{2} (x) = \frac{16}{25}$ $cos^2(x)=16/25$

$cos (x) = \pm \frac{4}{5}$ $cos(x) = pm 4/5$

Since $0 < x < 360^{\circ}$ $0< x < 360^circ$ , we can't rule out either value of $cos (x)$ $cos(x)$ .