|=0 [Applying R2→R2−R1 and R3→R3−R1, we get (2‌cos‌x+sin‌x)|
1
cos‌x
cos‌x
0
sin‌x−cos‌x
0
0
0
sin‌x−cos‌x
|=0 ‌⇒(2‌cos‌x+sin‌x)[1⋅(sin‌−cos‌x)2]=0 ‌⇒(2‌cos‌x+sin‌x)(sin‌x−cos‌x)2=0 ‌⇒2‌cos‌x=−sin‌x‌ or ‌sin‌x=cos‌x ‌⇒tan‌x=−2‌ which is not possible as for ‌ ‌−‌
Ï€
4
≤x≤‌
Ï€
4
,‌ we get ‌−1≤tan‌x≤1 ‌‌ or ‌‌‌tan‌x=1 So, x=‌