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Question

Question: How do you solve \( \sin (4x) = 1 \) ?...

How do you solve sin(4x)=1\sin (4x) = 1 ?

Explanation

Solution

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\sin (4x) = 1
Then next we will take the inverse of the sine function.
sin(4x)=1 4x=sin1(1)   \sin (4x) = 1 \\\ 4x = {\sin ^{ - 1}}(1) \;
Now we know that the value of sin900\sin {90^0} is 11 . Hence, now we substitute the value of sin900\sin {90^0} and hence evaluate the value of xx .
4x=sin1(1) 4x=π2 x=π8   4x = {\sin ^{ - 1}}(1) \\\ 4x = \dfrac{\pi }{2} \\\ \,\,x = \dfrac{\pi }{8} \;
Hence, the value of xx is π8\dfrac{\pi }{8} .
So, the correct answer is “ π8\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 sin900=1\sin {90^0} = 1 and sin1(1)=900=π2{\sin ^{ - 1}}(1) = {90^0} = \dfrac{\pi }{2} .