Question

If cos 2x + 2 cos x = 1 then, $(2-co{s}^{2}x)si{n}^{2}x$ is equal to

• A. 1
• B. -1
• C. $-\sqrt{5}$
• D. $\sqrt{5}$

Answer 1

The correct option is A

1

We have,

cos 2x + 2 cos x = 1

$\Rightarrow 2co{s}^2$ x - 1 + 2 cos x = 1

$\Rightarrow 2co{s}^2$ x + 2 cos x - 2 = 0

$\Rightarrow 2co{s}^2$x + cos x - 1 = 0

$\Rightarrow cosx=\frac{-1\pm \sqrt{1^2+4}}{2}\Rightarrow cosx=\frac{-1\pm \sqrt{5}}{2}\Rightarrow cosx=\frac{-1+\sqrt{5}}{2}$

Now,

$(2-co{s}^{2}x)si{n}^{2}x$

$=[2-{\left(\frac{-1\pm \sqrt{5}}{2}\right)}^2](1-co{s}^{2}x)=[2-\frac{1}{4}(1-2\sqrt{5}+5\left)\right](1-\frac{1}{4}(1-2\sqrt{5}+5\left)\right)=\frac{1}{4}(1+\sqrt{5})(\sqrt{5}-1)=\frac{4}{4}=1$