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Find the integrals of the functions cos2xcos4xcos6x View Solution Prove that: 1+cos2x+cos4x+cos6x=4cosxcos2xcos3x View Solution =(3)/(5)lim_(x rarr0)((1-cos2x*cos4x*cos6x*cos8x*cos10x)/(x^(2))) View Solution cos2x-cos8x+cos6x=1 View Solution Find the sum to n terms ...
4.化简:1+cos2x+cos4x+cos6x= 相关知识点: 试题来源: 解析 4.化简:1+cos2x+cos4x+cos6x=4 cos r cos 2r cos3r 结果一 题目 4.化简:1+cos2x+cos4x+cos6x= 答案 4. 4cosxcos2xcos3x相关推荐 14.化简:1+cos2x+cos4x+cos6x= 反馈 收藏 ...
cos4x+1=2cos²2x 由cos2x+cos4x+1＝cos6x得 2cos²2x+cos2x=cos4xcos2x-sin4xsin2x 2cos²2x+cos2x(1-cos4x)=-sin4xsin2x 2cos²2x+2cos2xsin²2x=-sin4xsin2x 2cos²2x=-4cos2xsin²2x cos2x(cos2x+2sin²2x)=0 因为 cos2x+2si...
cos2x+cos4x=1+cos6x 2cos3xcosx=2cos23x cos3x(cosx−cos3x)=0, ①cos3x=0时,3x=π2+kπ, x=π6+kπ3,k∈Z, ∴x=π6,π2,5π6,7π6,3π2,11π6. ②cosx−cos3x=0 时,3x−x=2kπ,k∈Z或3x+x=2kπ,k∈Z, x=kπ,k∈Z 或x=kπ2,k∈Z, x=π,0 或x=π2,π,...
解：1-cosxcos(2x)cos(3x)=1-½[cos(2x+x)+cos(2x-x)]cos(3x)=1-½[cos(3x)+cosx]cos(3x)=1-½cos²(3x)-½cosxcos(3x)=1-¼[1+cos(6x)]-¼[cos(3x+x)+cos(3x-x)]=¾-¼cos(6x)-¼[cos(4x)+cos(2x)]=&#...
怎么可能,有问题的式子怎么证明相等啊,不信我举反例就可推翻了 x=30度时=cos60+cos120+cos180=0.5-0.5-1=-1 x=60度时=cos120+cos240+cos360=-0.5-0.5+1=0
解：1-cosxcos(2x)cos(3x)=1-½[cos(2x+x)+cos(2x-x)]cos(3x)=1-½[cos(3x)+cosx]cos(3x)=1-½cos²(3x)-½cosxcos(3x)=1-¼[1+cos(6x)]-¼[cos(3x+x)+cos(3x-x)]=¾-¼cos(6x)-¼[cos(4x)+cos(2x)]=&#...
代数输入 三角输入 微积分输入 矩阵输入 cos(6x)=32cos(x)6−48cos(x)4+18cos(x)2−1 求解x 的值 (复数求解) x∈C 求解x 的值 x∈R 图表
解答一 举报 各项降次后方程变形为1+cos2x+cos4x+cos6x=0(1+cos6x)+(cos2x+cos4x)=02cos²3x+2cos3xcosx=0cos3x(cos3x+cosx)=02cos3xcos2xcosx=0cos3x=0 或 cos2x=0 或 cosx=0x=kπ/3±π/6,kπ/2±π4,kπ±π/2 (k∈Z) 解析看不懂?免费查看同类题视频解析查看解答 ...