Đáp án + Giải thích các bước giải:
`1.`
`sin ( 3x - (5π)/6 ) + cos ( 3x + (3π)/4 ) = 0`
`<=> cos ( 3x + (3π)/4 ) = -sin ( 3x - (5π)/6 ) `
`<=> cos ( 3x + (3π)/4 ) = sin ( 3x - (5π)/6 +pi) `
`<=> cos ( 3x + (3π)/4 ) = sin ( 3x + pi/6) `
`<=> cos ( 3x + (3π)/4 ) = cos ( 3x + pi/6-pi/2) `
`<=> cos ( 3x + (3π)/4 ) = cos ( 3x-pi/3) `
`<=>`\(\left[ \begin{array}{l}3x+\dfrac{3\pi}{4}=3x-\dfrac{\pi}{3}+k2\pi\\3x+\dfrac{3\pi}{4}=-3x+\dfrac{\pi}{3}+k2\pi\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}0x=-\dfrac{13\pi}{12}+k2\pi(VN)\\6x=-\dfrac{5\pi}{12}+k2\pi\end{array} \right.\)
`<=>x=-(5pi)/72+(kpi)/3(kinZZ)`
`2.`
`cos(x/2+π/4)=0`
`<=>x/2+pi/4=pi/2+kpi`
`<=>x/2=pi/4+kpi`
`<=>x=pi/2+k2pi(kinZZ)`
Do `x in(pi;8pi)=>pi<pi/2+k2pi<8pi`
`<=>1/4<k<15/4=>k=1;2;3`
`=>x=(5pi)/2,x=(9pi)/2,x=(13pi)/2`
Vậy có 3 nghiệm thỏa mãn.
