+8 Given \(sin(2x) =-\frac{\sqrt{15}}{8} \) and \(tan(2x) > 0, 0 \leq x \leq 2\pi\), find \(cos(x)\)
+399 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
+399 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
