Question

The total number of solutions of cosx = $\sqrt{1-sin2x}$ in (0, 2$\pi$ ) is equal to

• A. 2
• B. 3
• C. 5
• D. 1

Answer 1

The correct option is A

2

cos x = $\sqrt{1-sin2x}$ = $\sqrt{co{s}^{2}x+si{n}^{2}x-2sinxcosx}$

= |sinx - cosx|

Case1: sinx ≤ cosx

⇒ cosx = cosx - sinx ⇒ sinx = 0

Where x ∈ (0, $\frac{\pi }{4}$) ∪ ( $\frac{5\pi }{4}$, 2$\pi$)

⇒x = 2$\pi$ neglecting x = $\pi$

Case 2: sinx > cosx

⇒ tanx = 2 where x ∈ ( $\frac{\pi }{4}$, $\frac{5\pi }{4}$)

∴ x = $ta{n}^{-1}$(2). Hence two solutions