tan7x=-tan3x ==>tan7x=tan(π-3x)==>7x=kπ+π-3x==>10x=(k+1)π==>x=(k+1)π/10,k∈Zcos5x=-cosx ==>cos5x=cos(π-x)==>5x=2kπ±(π-x)==>5x±x=(2k±1)π==>x=(2k±1)π/6或x=(2k±1)π/4,k∈Zx=kπ/10可以吗可以 更简