Question

Solve the equation: $cos^{-1}(\sqrt{3}.x) + cos^{-1}x = \frac{\pi}{2}$.

Answer 1

Answer:

$x_{1}=\frac{1}{2} \\\\x_{2}=-\frac{1}{2}$

Step-by-step explanation:

As we know that

$cos^{-1}(x)+cos^{-1}y=cos^{-1}(xy-\sqrt{1-x^{2} } \sqrt{1-y^{2}\:)$

$cos^{-1}(\sqrt{3} x) + cos^{-1}x=cos^{-1}(\sqrt{3} x^{2}-\sqrt{1-(\sqrt{3} x)^{2} } \sqrt{1-x^{2}\:)$

$=> cos^{-1}(\sqrt{3} x^{2}-\sqrt{1-3x^{2} } \sqrt{1-x^{2}}}\:)=\frac{\pi }{2}$

taking cos both sides

$cos[cos^{-1}(\sqrt{3} x^{2}-\sqrt{1-3x^{2} } \sqrt{1-x^{2}}}\:)]=cos[\frac{\pi }{2}]$

as we know that cos 90°=0

$(\sqrt{3} x^{2}-\sqrt{1-3x^{2} } \sqrt{1-x^{2}}})=0\\\\=>\sqrt{3} x^{2}=\sqrt{1-3x^{2} } \sqrt{1-x^{2}}}$

taking square both sides

$3x^{4}=({1-3x^{2} )({1-x^{2})\\\\\\=> 3x^{4}=1-x^{2}-3x^{2}+3x^{4}\\\\\\=>1-4x^{2}=0$

$x^{2}=\frac{1}{4}\\ \\\\x=\sqrt{\frac{1}{4} } \\\\\\x_{1}=\frac{1}{2} \\\\x_{2}=-\frac{1}{2}$

Answer 2

it is given that,
$cos^{-1}(\sqrt{3}x)+cos^{-1}x=\frac{\pi}{2}$

Let $cos^{-1}(\sqrt{3}x)=A$
cosA = √3x/1 = b/h
so, b = √3x , h = 1
then, p = √(h² - b²) = √(1 - 3x²)
so, sinx = √(1 - 3x²).........(1)

similarly, $cos^{-1}x=B$
cosB = x
then, sinB = √(1 - x²) .......(2)


then, A + B = π/2

now, cos(A + B) = cosπ/2

or, cosA.cosB - sinA.sinB = 0

or, √3x. x - √(1 - 3x²) √(1 - x²) = 0

or, √3 x² = √(1 - 3x²) √(1 - x²)

squaring both sides,

or, 3x⁴ = (1 - 3x²)(1 - x²)

or, 3x⁴ = 1 - x² - 3x² + 3x⁴

or, 1 - 4x² = 0

or, x² = 1/4

taking square root both sides,

x = ±1/2

hence, x = 1/2 and -1/2