Question

If 3 – 2 cos x – 4 sin x – cos 2x + sin 2x = 0, then x =( n $ϵ$ Z)

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

$3–2cosx–4sinx–cos2x+sin2x=0\Rightarrow 3–2cosx–4sinx–(2co{s}^{2}x–1)+2sinxcosx=0\Rightarrow 4–4sinx–2cosx–2co{s}^{2}x+2sinxcosx=0\Rightarrow 2–2sinx+sinxcosx–cosx–co{s}^{2}x=0\Rightarrow 2(1–sinx)–cosx(1–xinx)–(1–si{n}^{2}x)=0\Rightarrow (1–sinx)[2–cosx–1–sinx]=0\Rightarrow 1–sinx=0or1–cosx–sinx=0\Rightarrow sinx=1orcosx+sinx=1\Rightarrow sinx=1orcos(x-\frac{\pi }{4})=\frac{1}{\sqrt{2}}\Rightarrow x=n\pi +(-n{)}^{n}orx-\frac{\pi }{4}=2n\pi +\frac{\pi }{4}\Rightarrow x=n\pi +(-1{)}^n\frac{\pi }{2}or2n\pi +\frac{\pi }{2}$