Giải thích các bước giải:
`sinx+co sx=0`
`⇔sinx=-cosx=sin(x-(π)/(2))`
`⇔`\(\left[ \begin{array}{l}x=x-\dfrac{π}{2}+k2π\\x=π-x+\dfrac{π}{2}+k2π\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}0x=\dfrac{-π}{2}+k2π(loại)\\2x=\dfrac{3π}{2}+k2π\end{array} \right.\)
`<=>`$x=\dfrac{3π}{4}+kπ=π-\dfrac{π}{4}+kπ=\dfrac{-π}{4}+kπ(k∈Z)$
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`sinx+cosx=\sqrt{2}.sin(x+(π)/(4))=\sqrt{2}.cos(x-(π)/(4))`
Áp dụng:
`sinx+co x=0`
`⇔\sqrt{2}.sin(x+(π)/(4))=0`
`⇔sin(x+(π)/(4))=0`
`⇔x+(π)/(4)=kπ`
`⇔x=(-π)/(4)+kπ(k∈Z)`