log(2)tanx=1+log(2)sinx的解集为

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/12 10:17:40
log(2)tanx=1+log(2)sinx的解集为

log(2)tanx=1+log(2)sinx的解集为
log(2)tanx=1+log(2)sinx的解集为

log(2)tanx=1+log(2)sinx的解集为
log(2)tanx=1+log(2)sinx等价于
log(2)tanx=log(2)2+log(2)sinx
即log(2)tanx=log(2)(2sinx)
等价于tanx=2sinx>0等价于
sinx=2sinxcosx>0
即sinx(2cosx-1)=0且sinx>0
所以cosx=1/2,且x∈(2kπ,2kπ+π)
得x=2kπ+π/3 (k∈Z)
所以原方程的解集为
{x|x=2kπ+π/3(k∈Z)}

log(2)tanx=1+log(2)sinx
log(2)tanx=log(2)2+log(2)sinx
log(2)tanx=log(2)(2sinx)
tanx=2sinx
sinx=2sinxcosx
sinx(2cosx-1)=0
sinx=0或cosx=1/2
x=2kπ或2kπ+π或2kπ+π/3或2kπ-π/3 (k∈Z)
解集为
{x|x=2kπ或2kπ+π或2kπ+π/3或2kπ-π/3 (k∈Z)}

log(2)tanx=1+log(2)sinx
log(2)tanx=log(2)2+log(2)sinx
log(2)tanx=log(2)(2sinx)
tanx=2sinx
sinx=2sinxcosx
sinx(2cosx-1)=0
sinx=0或cosx=1/2
所以x=kπ或者x=2kπ+π/3或者x=2kπ-π/3 k为整数

log(2)tanx=1+log(2)sinx的解集为 log(2)tanx=1+log(2)sinx的解集为 请写出过程, 函数y=根号下log 1/2 tanx的定义域,(0.5是底数,tanx是真数) 求函数y=log(1/2)(tanx-根号3)定义域kπ+π/3 计算:1/[log(2)3]+1/[log(5)3]=? log(X+5)+log(X+2)=1 log(27)32 × log(64)27+log(9)2 × log(4)根号27=?log(27)32 × log(64)27+log(9)2 × log(4)根号27=? log(2)(25)×log(3)(1/16)×log(5)(1/9)=? log(3)4*log(4)8*log(8)M =log(2)16 M是多少呃.... 求大神指点,lingo出现错误代码11,这是为什么?model:!目标函数;min=(0.005*((@log(1+k))^2+(@log(1+2*k))^2+(@log(1+3*k))^2+(@log(1+4*k))^2+(@log(1+5*k))^2))!/(@log(1+k)+@log(1+2*k)+@log(1+3*k)+@log(1+4*k)+@log(1+5*k));!约束条件;(@log 求等式log(2)3*log(3)4*log(4)5*log(5)6*log(6)7*log(7)m=log(3)9时m的值? 若|log(sinx)cosx|+|log(cosx)根号tanx|=1,求锐角x.要具体过程. log(2x)+log(3y)-log(2z) 怎么化简?log3(x^2-2x-6)=2 log(3x+6)=1+log(x)求解,求每步详...log(2x)+log(3y)-log(2z) 怎么化简?log3(x^2-2x-6)=2 log(3x+6)=1+log(x)求解, 关于数学对数的换底公式推论的问题已知 log(2)(3) = a,log(3(7)=b,用a,b表示log(42)(56)因为log(2)(3)=a,则1/a=log(3)(2),又∵log(3)(7)=b,∴log(42)(56)=log(3)(56)/log(3)(42)=log(3)(7)+3·log(3)(2)/log(3)(7)+log(3)(2)+1=ab+3/ab+b+1 关于数学对数的换底公式推论的问题已知 log(2)(3) = a,log(3(7)=b,用a,b表示log(42)(56)因为log(2)(3)=a,则1/a=log(3)(2),又∵log(3)(7)=b,∴log(42)(56)=log(3)(56)/log(3)(42)=log(3)(7)+3·log(3)(2)/log(3)(7)+log(3)(2)+1=ab+3/ab+b+1 证明对数运算法则(1)log(a)(MN)=log(a)(M)+log(a)(N);   (2)log(a)(M/N)=log(a)(M)-log(a)(N);(1)log(a)(MN)=log(a)(M)+log(a)(N);   (2)log(a)(M/N)=log(a)(M)-log(a)(N);   (3)log(a)(M^n)=nlog(a)(M) (n∈R) 设 x=log11 ,y=log 12 ,z=log 13,令S={log 1,log 2,log 3,…,log 100} ,则S 中有多少个元素可表为ax+by+cz+d ( a ,b ,c ,d εQ ). log(x+5)=log(x)+log(2) log(8)+log(x)=log(24) 2log(x)=log(2x)+log(3)