
How do you solve $ \sin (4x) = 1 $ ?
Answer
523.5k+ views
Hint: Take out all the like terms to one side and all the alike terms to the other side. Take out all the common terms. Reduce the terms on the both sides until they cannot be reduced any further if possible. Then finally evaluate the value of the unknown variable.
Complete step by step solution:
First we will start off by taking all the like terms to one side.
$ \sin (4x) = 1 $
Then next we will take the inverse of the sine function.
$
\sin (4x) = 1 \\
4x = {\sin ^{ - 1}}(1) \;
$
Now we know that the value of $ \sin {90^0} $ is $ 1 $ . Hence, now we substitute the value of $ \sin {90^0} $ and hence evaluate the value of $ x $ .
$
4x = {\sin ^{ - 1}}(1) \\
4x = \dfrac{\pi }{2} \\
\,\,x = \dfrac{\pi }{8} \;
$
Hence, the value of $ x $ is $ \dfrac{\pi }{8} $ .
So, the correct answer is “ $ \dfrac{\pi }{8} $ ”.
Note: o cross multiply terms, you will multiply the numerator in the first fraction times the denominator in the second fraction, then you write that number down. Then you multiply the numerator of the second fraction times the number in the denominator of your first fraction, and then you write that number down. By Cross multiplication of fractions, we get to know if two fractions are equal or which one is greater. This is especially useful when you are working with larger fractions that you are not sure how to reduce. Cross multiplication also helps us to solve for unknown variables in fractions.
While cross multiplying the terms, multiply the terms step-by-step to avoid any mistakes. Always take the variables to one side and integer type of terms to the other side. Also remember that $ \sin {90^0} = 1 $ and $ {\sin ^{ - 1}}(1) = {90^0} = \dfrac{\pi }{2} $ .
Complete step by step solution:
First we will start off by taking all the like terms to one side.
$ \sin (4x) = 1 $
Then next we will take the inverse of the sine function.
$
\sin (4x) = 1 \\
4x = {\sin ^{ - 1}}(1) \;
$
Now we know that the value of $ \sin {90^0} $ is $ 1 $ . Hence, now we substitute the value of $ \sin {90^0} $ and hence evaluate the value of $ x $ .
$
4x = {\sin ^{ - 1}}(1) \\
4x = \dfrac{\pi }{2} \\
\,\,x = \dfrac{\pi }{8} \;
$
Hence, the value of $ x $ is $ \dfrac{\pi }{8} $ .
So, the correct answer is “ $ \dfrac{\pi }{8} $ ”.
Note: o cross multiply terms, you will multiply the numerator in the first fraction times the denominator in the second fraction, then you write that number down. Then you multiply the numerator of the second fraction times the number in the denominator of your first fraction, and then you write that number down. By Cross multiplication of fractions, we get to know if two fractions are equal or which one is greater. This is especially useful when you are working with larger fractions that you are not sure how to reduce. Cross multiplication also helps us to solve for unknown variables in fractions.
While cross multiplying the terms, multiply the terms step-by-step to avoid any mistakes. Always take the variables to one side and integer type of terms to the other side. Also remember that $ \sin {90^0} = 1 $ and $ {\sin ^{ - 1}}(1) = {90^0} = \dfrac{\pi }{2} $ .
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

