To find the values of \(\sin x\) given the equation \(8\sin x = 4 + \cos x\), we can use trigonometric identities to rewrite the equation in terms of a single trigonometric function.

The given equation is:
\[8\sin x = 4 + \cos x\]

We know the Pythagorean identity: \(\sin^2 x + \cos^2 x = 1\). Our goal is to express \(\cos x\) in terms of \(\sin x\) (or vice versa) to solve for \(\sin x\). Since the equation involves both \(\sin x\) and \(\cos x\), and we’re looking to find \(\sin x\), let’s isolate \(\cos x\) and then use the Pythagorean identity.

Rearrange the given equation to isolate \(\cos x\):
\[\cos x = 8\sin x – 4\]

Using the Pythagorean identity \(\cos^2 x = 1 – \sin^2 x\), we substitute for \(\cos x\) in the equation:
\[1 – \sin^2 x = (8\sin x – 4)^2\]

Expanding the right side and moving all terms to one side gives us a quadratic equation in \(\sin x\):
\[1 – \sin^2 x = 64\sin^2 x – 64\sin x + 16\]

Combine like terms:
\[65\sin^2 x – 64\sin x + 15 = 0\]

This is a quadratic equation in \(\sin x\). To solve for \(\sin x\), we use the quadratic formula where \(a = 65\), \(b = -64\), and \(c = 15\):
\[\sin x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\]
\[= \frac{64 \pm \sqrt{(-64)^2 – 4 \cdot 65 \cdot 15}}{2 \cdot 65}\]
\[= \frac{64 \pm \sqrt{4096 – 3900}}{130}\]
\[= \frac{64 \pm \sqrt{196}}{130}\]
\[= \frac{64 \pm 14}{130}\]

So, we have two possible solutions for \(\sin x\):
1. \(\sin x = \frac{64 + 14}{130} = \frac{78}{130} = \frac{39}{65} = \frac{3}{5}\)
2. \(\sin x = \frac{64 – 14}{130} = \frac{50}{130} = \frac{25}{65} = \frac{5}{13}\)

Therefore, the values of \(\sin x\) that satisfy the given equation are \(\frac{3}{5}\) and \(\frac{5}{13}\).

To find the values of \(\sin x\) given the equation \(8\sin x = 4 + \cos x\), we can use trigonometric identities to rewrite the equation in terms of a single trigonometric function.

The given equation is:
\[8\sin x = 4 + \cos x\]

We know the Pythagorean identity: \(\sin^2 x + \cos^2 x = 1\). Our goal is to express \(\cos x\) in terms of \(\sin x\) (or vice versa) to solve for \(\sin x\). Since the equation involves both \(\sin x\) and \(\cos x\), and we’re looking to find \(\sin x\), let’s isolate \(\cos x\) and then use the Pythagorean identity.

Rearrange the given equation to isolate \(\cos x\):
\[\cos x = 8\sin x – 4\]

Using the Pythagorean identity \(\cos^2 x = 1 – \sin^2 x\), we substitute for \(\cos x\) in the equation:
\[1 – \sin^2 x = (8\sin x – 4)^2\]

Expanding the right side and moving all terms to one side gives us a quadratic equation in \(\sin x\):
\[1 – \sin^2 x = 64\sin^2 x – 64\sin x + 16\]

Combine like terms:
\[65\sin^2 x – 64\sin x + 15 = 0\]

This is a quadratic equation in \(\sin x\). To solve for \(\sin x\), we use the quadratic formula where \(a = 65\), \(b = -64\), and \(c = 15\):
\[\sin x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\]
\[= \frac{64 \pm \sqrt{(-64)^2 – 4 \cdot 65 \cdot 15}}{2 \cdot 65}\]
\[= \frac{64 \pm \sqrt{4096 – 3900}}{130}\]
\[= \frac{64 \pm \sqrt{196}}{130}\]
\[= \frac{64 \pm 14}{130}\]

So, we have two possible solutions for \(\sin x\):
1. \(\sin x = \frac{64 + 14}{130} = \frac{78}{130} = \frac{39}{65}\)= \frac{3}{5}\)
2. \(\sin x = \frac{64 – 14}{130} = \frac{50}{130} = \frac{25}{65} = \frac{5}{13}\)

Therefore, the values of \(\sin x\) that satisfy the given equation are \(\frac{3}{5}\) and \(\frac{5}{13}\).

To find the value of the expression \(x^3 + 27x^2 + 243x + 631\) when \(x = 2\), we can substitute the value of \(x\) into the expression and simplify:

\[
\begin{aligned}
x^3 + 27x^2 + 243x + 631 &= 2^3 + 27(2)^2 + 243(2) + 631 \\
&= 8 + 27(4) + 486 + 631 \\
&= 8 + 108 + 486 + 631 \\
&= 1233
\end{aligned}
\]

So, the value of the expression when \(x = 2\) is \(1233\).

Understanding the Problem

The shopkeeper has 660 Kesar Petha and 510 Paan Petha. He wants to arrange them in stacks such that each stack has the same number of Pethas, and the arrangement takes the least area of the tray. We need to find the number of Pethas that can be placed in each stack for this purpose.

This problem essentially asks for the greatest common divisor (GCD) of the two numbers, which will give us the largest stack size that can be used to arrange both types of Petha in an optimal manner.

Solving the Problem

Let’s find the GCD of 660 and 510.

We can use the Euclidean algorithm to find the GCD:

1. Divide 660 by 510 and find the remainder: \(660 = 510 \times 1 + 150\)
2. Now, divide 510 by the remainder from the previous step: \(510 = 150 \times 3 + 60\)
3. Continue this process until the remainder is 0: \(150 = 60 \times 2 + 30\) and \(60 = 30 \times 2 + 0\)

The last non-zero remainder is 30, so the GCD of 660 and 510 is 30.

Conclusion

The number of Pethas that can be placed in each stack for this purpose is 30.

Understanding the Problem

In a partnership business, A and B invest in the ratio of 4:9. A withdraws his investment after 6 months. Despite A’s withdrawal, they still receive profits in the same ratio as their initial investments (4:9). We need to find out for how long B’s investment was used.

Solving the Problem

Let’s denote the period for which B’s investment was used as \(x\) months.

The profit earned by a partner in a business is directly proportional to the product of their investment and the time period for which the investment is made.

Given that A’s investment was used for 6 months and B’s investment was used for \(x\) months, we can write the following equation based on the proportion of their profits:

\[ \frac{\text{A’s investment} \times \text{A’s time}}{\text{B’s investment} \times \text{B’s time}} = \frac{4}{9} \]

Plugging in the values:

\[ \frac{4 \times 6}{9 \times x} = \frac{4}{9} \]

Simplifying:

\[ \frac{24}{9x} = \frac{4}{9} \]

\[ 24 = 4x \]

\[ x = \frac{24}{4} = 6 \]

Conclusion

B’s investment was used for 6 months.