cos(π/4+x)=3/5,17π/12

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cos(π/4+x)=3/5,17π/12
cos(π/4+x)=3/5,17π/12

cos(π/4+x)=3/5,17π/12
cos(π/4+x)
=cosπ/4cosx-sinπ/4sinx
=√2/2(cosx-sinx)
=3/5
cosx-sinx=3√2/5
sinx=cosx-3√2/5
两边平方
sin²x=1-cos²x=cos²x-6√2/5*cosx+18/25
2cos²x-6√2/5*cosx-7/25=0
50cos²x-30√2cosx-7=0
cosx=(3√2±4√2)/10
cosx=-√2/10或7√2/10
17π/12

虽然1楼回答得很好，但是我有更好的方法
∵x=(π/4+x)-π/4
5/3π∴sin(π/4+x)=-4/5
∴cosx=cos[(π/4+x)-π/4]=cos(π/4+x)cosπ/4+sin(π/4+x)sinπ/4[这一步用了两角和的三角函数公式tan(a+b)=(tana+tanb)/(1-tanatanb）]
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cos(π/4+x)=3/5,17π/12 cos(π/4+X)=3/5,17π/12 cos(π/4+x)=3/5,17π/12 cos(π/4+x)=3/5,17π/12 cos(π/4+x)=3/5,17/12π cos(π/4+x)=3/5,17π/12 cos(π/4+x)=3/5,17π/12 cos(π/4+x)=3/5,-17π/12 cos(2x-π)/[根号2cos(x+π/4)]=-1/5,0 已知sin[a-b]cos a-cos[b-a]sin a=3/5,b是第三象限角,求sin[b+5π/4]的值第一题1/2cos x-√3/2sin x第二题√3sin x+cos x第三题√2[sin x-cos x]第四题√2cos x-√6sin x 已知cos（x+45º)=3／5且17π/12＜x＜7π/4,求(sin2x+2cos²x)／1-tanx的值注意!括号里加的是cos^2x y=cos(π/3-x)cos[π/2（x-1)]判断奇偶性y=cos(π/3-x)cos(π/3+x)判断奇偶性三角函数解方程 (cos 3x)^3+ (cos 5x)^3 = 8 (cos 4x)^3 (cos x)^3 已知cos(π/4+x)=3/5 ,0 化简sin(3π-x)cos(x-3/2)cos(4π-x)/tan(x-5π)cos(π/2+x)sin(x-5/2π) 化简sin(3π-x)cos(x-3/2)cos(4π-x)/tan(x-5π)cos(π/2+x)sin(x-5/2π) 为什么cos(2x-π/3)-sin(5π/6) =cos(2x-π/3)? 三角函数运算为什么 cos2x+cos( 4π/3 -2x)=cos( π/3 +2x)