Question

sinx+cosx+3sinxcosx-1=0

 

Answer 1

Phương trình tương đương:

⇔(2sin⁡x−1)(cos⁡x+sin⁡x+2)=0

⇔cos⁡x(2sin⁡x−1)+sin⁡x(2sin⁡x−1)+2(2sin⁡x−1)=012sin⁡x+cos⁡x=−2sin⁡x+cos⁡x=2.sin⁡(x+π4)∈[−2;2]⇒sin⁡x+cos⁡x>−2" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-table; float:none; font-family:times new roman,times,serif; font-size:18.18px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">sin⁡x+cos⁡x=2.sin⁡(x+π4)∈[−2;2]⇒sin⁡x+cos⁡x>−2" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-table; float:none; font-family:times new roman,times,serif; font-size:18.18px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">sin⁡x+cos⁡x=2.sin⁡(x+π4)∈[−2;2]⇒sin⁡x+cos⁡x>−2 Do đó:

1)⇔sinx=12⇒⎡⎢ ⎢⎣π