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解三角形高考必考吗_解三角形高考
tamoadmin 2024-05-21 人已围观
简介1.2sin^2x+6sinxcosx-4=0; 2-sin^2x=3sinxcosx; 4-4sin^2x+sin^4x=9sin^2xcos^2x=9sin^2x-9sin^4x; 10sin^4x-13sin^2x+4=0; sin^2x=1/2或sin^2x=4/5; sinx=2/2或sinx=25/5; x=π/4+kπ/2或arctan(2)+kπ。2.(1/2)*sinx-(3/2
1.2sin^2x+6sinxcosx-4=0; 2-sin^2x=3sinxcosx; 4-4sin^2x+sin^4x=9sin^2xcos^2x=9sin^2x-9sin^4x; 10sin^4x-13sin^2x+4=0; sin^2x=1/2或sin^2x=4/5; sinx=±√2/2或sinx=±2√5/5; x=π/4+kπ/2或arctan(±2)+kπ。
2.(1/2)*sinx-(√3/2)*cosx=√2/2; sinxcos(π/3)-sin(π/3)cosx=sin(x-π/3)=√2/2; x-π/3=π/4+2kπ或3π/4+2kπ; x=7π/12+2kπ或13π/12+2kπ。
3.sin^25x=cos^2x;1/2-cos10x/2=cos2x/2+1/2; cos2x=-cos10x; 2x=10x+kπ; x=kπ/8。
4.(√2/2)*sinx+(√2/2)*cosx=sin(x+π/4)=1/2; x+π/4=π/6+2kπ或5π/6+2kπ; x=-π/12+2kπ或7π/12+2kπ; 在[0,4π]上的解为7π/12、23π/12、31π/12、47π/12; 和为9π。
5.(√2/2)*sinx+(√2/2)*cosx=sin(x+π/4)=(√2/2)*a; -1<=(√2/2)*a<=1; a∈[-1,1]。
6.(1/2)*sinx+(√3/2)*cosx=-a/2; sin(x+π/3)=-a/2; x=arcsin(-a/2)-π/3+2kπ或arcsin(-a/2)+2π/3+2kπ; x在[0,2π]上只有两个不同解，因此可知sin(x+π/3)不能为0、1、-1，即a的范围为a≠ 0且a≠ -2且a≠2。
