当cos2x＝SQR（2）÷3时,sinx的四次方＋cosx的四次方的值是

sin^2 x+cos^2 x=1,所以(sin^2 x+cos^2 x)^2=1 sin^4 x+cos^4 x+2cos^2 xsin^2 x=1 cos2x＝√2/3,即 2cos^2 x-1=√2/3 所以cos^2 x=(3+√2)/6,sin^2 x=1-cos^2 x=(3-√2)/6 sin^4 x+cos^4 x+2cos^2 xsin^2 x=sin^4 x+cos^4 x+2 (3+√2)/6 *(3-√2)/6 =sin^4 x+cos^4 +7/18=1 所以sin^4 x+cos^4 x=1-7/18=11/18