Question

What are the solutions to the equation sin²(θ) - 3sin(θ) - 4 = 0?

a) sin(θ) = 4, sin(θ) = -1
b) sin(θ) = 4, sin(θ) = -4
c) sin(θ) = 4, sin(θ) = 1
d) sin(θ) = 4, sin(θ) = 2

Answer 1

The solutions to the quadratic equation sin²(θ) - 3sin(θ) - 4 = 0 are found by factoring it into (sin(θ) - 4)(sin(θ) + 1) = 0, which gives sin(θ) = 4 and sin(θ) = -1. However, as the sine function has a range from -1 to 1, the only valid solution is sin(θ) = -1.

Step-by-step explanation:

The equation given is sin²(θ) - 3sin(θ) - 4 = 0, which is a quadratic equation in terms of sin(θ). We can solve this equation as if it were any other quadratic equation.

To solve it, let's substitute sin(θ) with a variable, say 'x'. This gives us: x² - 3x - 4 = 0. Factoring this equation we get:

(x - 4)(x + 1) = 0

So the solutions for 'x' are:

• x = 4
• x = -1

However, since 'x' represents sin(θ), and the sine of an angle cannot be greater than 1 or less than -1 in real numbers, the solution sin(θ) = 4 is not possible. Therefore, the only valid solution is sin(θ) = -1.

The correct answer to the given equation is thus:

• sin(θ) = -1