|=0 Taking (2cosx+sinx) common from C1, we get (2cosx+sinx)|
1
cosx
cosx
1
sinx
cosx
1
cosx
sinx
|=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 Expanding along C1, we get (2cosx+sinx)[1(sinx−cosx)2]=0 ⇒ (2cosx+sinx)(sinx−cosx)2=0 Now, if 2cosx+sinx=0, then 2cosx=−sinx ⇒tanx=−2 But here −
π
4
≤x≤
π
4
, we have −1≤tanx≤1, so no solution possible. or if (sinx−cosx)2=0, then sinx=cosx ⇒tanx=1=tan