Question

Solve the equation: $cos^{-1}x + sin^{-1}(\frac{x}{2}) = \frac{\pi}{6}$

Answer 1

Answer:

Step-by-step explanation:

As we know that

$sin^{-1}(x) =cos^{-1}(\sqrt{1-x^{2}} ) \\\\sin^{-1}(\frac{x}{2} ) =cos^{-1}(\sqrt{1-(\frac{x}{2})^{2}} ) \\\\\\=cos^{-1}(\sqrt{(\frac{4-x^{2}}{4})} ) \\\\=cos^{-1}\frac{\sqrt{4-x^{2}} }{2}$

now apply

$cos^{-1}x+cos^{-1}y=cos^{-1}[xy-\sqrt{1-x^{2}} \sqrt{1-y^{2}} ]\\\\cos^{-1}x+cos^{-1}\frac{\sqrt{4-x^{2}} }{2}=cos^{-1}[x\frac{\sqrt{4-x^{2}} }{2}-\sqrt{1-x^{2}} \sqrt{1-(\frac{\sqrt{4-x^{2}} }{2})^{2}} ]\\\\\\=cos^{-1}[x\frac{\sqrt{4-x^{2}} }{2}-\sqrt{1-x^{2}} \sqrt{1-(\frac{{4-x^{2}} }{4})} ]$

now taking cos both sides

$cos[cos^{-1}[x\frac{\sqrt{4-x^{2}} }{2}-\sqrt{1-x^{2}} \sqrt{1-(\frac{{4-x^{2}} }{4})} ]]=cos\frac{\pi}{6}$

$[x\frac{\sqrt{4-x^{2}} }{2}-\sqrt{1-x^{2}} \sqrt{1-(\frac{{4-x^{2}} }{4})} ]]=\frac{\sqrt{3}}{2}$

$x\frac{\sqrt{4-x^{2}} }{2}-\frac{\sqrt{3} }{2}=\sqrt{1-x^{2}}\sqrt{1-(\frac{{4-x^{2}} }{4})}$

taking square both sides

$\frac{x^{2}{(4-x^{2}} )}{4}-\frac{{3} }{4}-x\sqrt{4-x^{2} }=({1-x^{2}}) ({1-(\frac{{4-x^{2}} }{4})})\\\\4x^{2}-x^{4}-3-4x\sqrt{4-x^{2} }=(1-x^{2})(x^{2})\\\\4x^{2}-x^{4}-3-4x\sqrt{4-x^{2} }=(x^{2}-x^{4})\\\\3x^{2}-4x\sqrt{4-x^{2} }-3=0\\\\(3x^{2}-3)^{2}=(4x\sqrt{4-x^{2} })^{2}\\\\9x^{4}-18x^{2} +9=16x^{2} (4-x^{2} )\\\\=> 9x^{4}-18x^{2} +9-64x^{2} +16x^{4} =0\\\\25x^{4} -82x^{2} +9=0$

now solve this by quadratic formula as

$x_{1,2}=\frac{-b+-\sqrt{b^{2} -4ac} }{2a} \\\\x_{1,2}=\frac{82+-\sqrt{(82)^{2} -4(25)9} }{2(25)} \\\\\\=\frac{82+-\sqrt{5824} }{50}$

${x^{2}}_{1,2}=\frac{82+-8\sqrt{91} }{50}\\\\\\=\frac{41+-4\sqrt{91} }{25}$