2 câu trả lời
sin(4x+1)=13⇔[4x+1=arcsin13+k2π4x+1=π−arcsin13+k2π⇔[4x=arcsin13−1+k2π4x=π−arcsin13−1+k2π⇔[x=14(arcsin13−1+k2π)x=14(π−arcsin13−1+k2π).
Lời giải: sin(4x+1)=13⇔[4x+1=arcsin13+k2π4x+1=π−arcsin13+k2π⇔[4x=−1+arcsin13+k2π4x=−1+π−arcsin13+k2π⇔[x=−14+14arcsin13+kπ2x=−14+π4−14arcsin13+kπ2(k∈Z)