+0

# Urgent triq question

0
14
1
+8

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

Mar 3, 2024

#1
+394
+2

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

Mar 4, 2024

#1
+394
+2

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