Question

Solve. sqrt6*[sin^2(x)-cos^2(x)] , sin2x=sqrt3/3 ,
-3PI/4 14 years ago

Answer 1

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

Answer 2

Just to separate the conditions.

Answer 3

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

Answer 4

\[\sin2x=_{3}^{\sqrt{3}}\]

Answer 5

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

Answer 6

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.

Answer 7

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

Answer 8

Yes. :)

Answer 9

draw a diagram and find the side marked a



\[x = \sqrt{6}\]

\[\cos(2x) = \sqrt{6}/3\]

so the problem is

\[\sqrt{6}((\sqrt{3}/3)^2 - (\sqrt{6}/3)^2) = - 3\sqrt{6}/9 = \sqrt{6}/3\]