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рем 809 * x7 500 * # 809. |
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Answer» Cosxcos2xcos3x=or, (2cosxcos3x)(2cos2x)=1or, (cos4x+cos2x)(2cos2x)=1or, 2cos4xcos2x+(2cos²2x-1)=0or, 2cos4xcos2x+cos4x=0or, cos4x(2cos2x+1)=0Either, cos4x=0or, 4x=(2n+1)π/2or, x=(2n+1)π/8Or, 2cos2x+1=0or, 2cos2x=-1or, cos2x=-1/2or, cos2x=cos(-π/3)or, cos2x=cosπ/3 [neglecting the negative sign for cos]or, 2x=2nπ⁺₋π/3or, x=nπ⁺₋π/6(⁺₋ means plus-minus) |
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