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Given $sin(2x) =-\frac{\sqrt{15}}{8}$ and $tan(2x) > 0, 0 \leq x \leq 2\pi$ , find $cos(x)$

albertthesav Mar 3, 2024

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Best Answer

We know $\sin(2x)>0, \tan(2x)<0$ , therefore x is in quadrant 3. So $\cos(2x)$ must be negative.

Therefore, ${\cos}^{2}(2x)=1-{\sin}^{2}(2x)=\frac{49}{64}$ , $\cos(2x)=-\frac{7}{8}$ .

Becuase cos(2x) is in the third quadrant, then we know x must be in the second quadrant, so cos must be negative

Using our half-angle formula for cos:

$\cos(x)=-\sqrt{\frac{\frac{1}{8}}{2}}=-\frac{1}{4}$ . Therefore cos(x) = -1/4

hairyberry Mar 4, 2024

hairyberry · Answer 1 · 2024-03-04T01:16:44

We know $\sin(2x)>0, \tan(2x)<0$ , therefore x is in quadrant 3. So $\cos(2x)$ must be negative.

Therefore, ${\cos}^{2}(2x)=1-{\sin}^{2}(2x)=\frac{49}{64}$ , $\cos(2x)=-\frac{7}{8}$ .

Becuase cos(2x) is in the third quadrant, then we know x must be in the second quadrant, so cos must be negative

Using our half-angle formula for cos:

$\cos(x)=-\sqrt{\frac{\frac{1}{8}}{2}}=-\frac{1}{4}$ . Therefore cos(x) = -1/4