Solve Equations with Inverse Sine

Simplify the equation: 2sin(x) − 1 = 0 ⇒ sin(x) = 1/2. The inverse sine of 1/2 gives x = π/6. This value lies within the arcsin range.

sin(π/4) = √2/2. For a negative sine value, take the negative of the corresponding positive angle. Thus arcsin(−√2/2) = −π/4.

Since sin(π/3) = √3/2, the angle x = π/3 gives the required sine value. It’s also within the arcsin range, so x = π/3.

sin(π/4) = √2/2, and π/4 lies within the arcsin range. So θ = arcsin(√2/2) = π/4.

The sine and arcsine functions are inverses. That means if x is between −1 and 1, applying sine to arcsin(x) returns x itself. So sin(arcsin(x)) = x always holds true.

We are looking for the principal value, which is the angle between −90° and 90° whose sine equals 0.92.

Since 0.92 is close to 1, the angle must be large and close to 90°, because sine increases as the angle approaches 90°.

We also know some benchmark values:

sin(60°) ≈ 0.866

sin(90°) = 1

Since 0.92 is between 0.866 and 1, the angle must be between 60° and 90°, and closer to 90° than to 60°.

Of the choices, 66.8° is the only angle in that reasonable range.

Therefore, the principal value is 66.8°.

Apply sine to both sides: sin(arcsin(y)) = sin(−π/6). Since sin(−π/6) = −1/2, we get y = −1/2.

sin(π/6) = 1/2, and π/6 is in the arcsin range [−π/2, π/2]. Thus, the correct principal angle is π/6.

Start with the equation:

3sin(x) + 1 = 0

3sin(x) = −1

sin(x) = −1/3

Now we want the principal value of x.

The arcsin function returns angles only in the interval:

−π/2 ≤ x ≤ π/2

Because −1/3 is negative, the principal value must be a negative angle in that interval.

The direct inverse-sine expression for the principal value is:

x = arcsin(−1/3)

sin(−π/2) = −1. Thus, arcsin(−1) = −π/2 since that’s in the principal range.

arcsin(x) only accepts inputs between −1 and 1. The output (angle) lies between −π/2 and π/2. That’s the definition of the function’s domain and range.

We know sin(π/6) = 1/2. The principal range for arcsin is [−π/2, π/2], and π/6 lies in this range. Therefore, x = arcsin(1/2) = π/6.

Because the sine value is negative, the angle must lie below the x-axis. So θ = −arcsin(3/5). This is within the principal range of arcsin.

First, 2x = arcsin(1/2) = π/6. Divide both sides by 2 → x = π/12. This gives the principal solution.

The sine of −π/2 equals −1. Because −π/2 is included in the range [−π/2, π/2], arcsin(−1) = −π/2.

Because the sine value is negative, the angle must be negative. The principal solution is x = −arcsin(0.8). This falls within the valid arcsin range.

Sine and arcsine are inverse operations. So sin(arcsin(x)) = x. Therefore, sin(arcsin(−7/10)) = −7/10.

The sine of an angle equals rise/hypotenuse. So θ = arcsin(12/30). This expression gives the correct relationship between rise and hypotenuse.

Using a calculator: arcsin(2/3) ≈ 0.7297 radians. Rounded to two decimals, x = 0.73 radians. This is the principal value.

The sine of 0 equals 0. Therefore, arcsin(0) = 0 since 0 is in the range [−π/2, π/2].