Solve. sqrt6*[sin^2(x)-cos^2(x)] , sin2x=sqrt3/3 ,
-3PI/4

Question

Solve. sqrt6*[sin^2(x)-cos^2(x)] , sin2x=sqrt3/3 ,
-3PI/4<x<-PI/2.

dumbcow · Answer

i don't get the equation, why is there a comma?

anonymous · Answer

Just to separate the conditions.

anonymous · Answer

$$\sqrt{6}*(\sin^2x-\cos^2x) .    $$

anonymous · Answer

$$\sin2x=_{3}^{\sqrt{3}}$$

dumbcow · Answer

thats not an equation though
however, sin^ -cos^2 = -cos(2x)

anonymous · Answer

That I know. But I need to get sin or cos from the sin2x and then plug them into the first equation to get the value of x.

dumbcow · Answer

ok gotcha, so you need to evaluate the expression given those conditions :)

anonymous · Answer

Yes. :)

campbell_st · Answer

draw a diagram   and find the side marked a

|dw:1334994493878:dw|

$$x = \sqrt{6}$$

$$\cos(2x) = \sqrt{6}/3$$

so the problem is 

$$\sqrt{6}((\sqrt{3}/3)^2 - (\sqrt{6}/3)^2) = - 3\sqrt{6}/9 = \sqrt{6}/3$$