Solve for x - sin -1 6 x -sin -1 6 -3 x -2

X:sin^ 16x+sin^ 16√(3)x= π/2⇒ sin^ 16x= π/2 sin^ 16√(3)x ⇒ sinsin^ 16x=sin ( π/2 sin^ 16√(3)x )⇒ 6x= sin ( π/2+sin^ 16√(3)x ) ⇒ 6x= cos(sin^ 16√(3)x)= √(1 10 ...

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Byju's Answer

Standard XII

Mathematics

Domain and Range of Basic Inverse Trigonometric Functions

Solve for x :...

Question

Solve for $x : s i n^{- 1} 6 x + s i n^{- 1} 6 \sqrt{3} x = - \frac{π}{2}$

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Solution

$x : s i n^{- 1} 6 x + s i n^{- 1} 6 \sqrt{3} x = - \frac{π}{2} \Rightarrow s i n^{- 1} 6 x = - \frac{π}{2} - s i n^{- 1} 6 \sqrt{3} x$
$\Rightarrow s i n s i n^{- 1} 6 x = s i n (- \frac{π}{2} - s i n^{- 1} 6 \sqrt{3} x) \Rightarrow 6 x = - s i n (\frac{π}{2} + s i n^{- 1} 6 \sqrt{3} x)$
$\Rightarrow 6 x = - c o s (s i n^{- 1} 6 \sqrt{3} x) = - \sqrt{1 - 108 x^{2}}$
$\Rightarrow 36 x^{2} = 1 - 108 x^{2} \Rightarrow 144 x^{2} = 1 \Rightarrow x = \pm \frac{1}{12}$

Since $x = \frac{1}{12}$ does not satisfy the given equation, $∴ x = - \frac{1}{12}$

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Similar questions

Q. Solve the equation

{s i n}^{- 1} 6 x + {s i n}^{- 1} 6 \sqrt{3} x = \frac{- π}{2}

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x : {sin}^{- 1} \frac{5}{x} + {sin}^{- 1} \frac{12}{x} = \frac{π}{2}, x \neq 0

s i n^{- 1} 6 x + s i n^{- 1} 6 \sqrt{3} x = - π / 2

if x is equal to

Q. If

S i n^{- 1} (6 x) + S i n^{- 1} (6 \sqrt{3} x) = - \frac{π}{2}

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s i n^{- 1} 6 x + s i n^{- 1} 6 \sqrt{3} x = - \frac{π}{2}

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Solve for x:sin−16x+sin−16√3x=−π2

x:sin−16x+sin−16√3x=−π2⇒sin−16x=−π2−sin−16√3x ⇒sinsin−16x=sin(−π2−sin−16√3x)⇒6x=−sin(π2+sin−16√3x) ⇒6x=−cos(sin−16√3x)=−√1−108x2 ⇒36x2=1−108x2⇒144x2=1⇒x=±112 Since x=112 does not satisfy the given equation,∴ x=−112

Solve for $x : s i n^{- 1} 6 x + s i n^{- 1} 6 \sqrt{3} x = - \frac{π}{2}$