Questions: In the triangle above, the sine of x^0 is 0.6. What is the cosine of y^∘? For what real value of x is the equation above true?

Transcript text: In the triangle above, the sine of $x^{0}$ is 0,6 . What is the cosine of $y^{\circ}$ ? For what real value of $x$ is the equation above true?

Solution

Solution Steps

Step 1: Relate sine and cosine

In a right triangle, the sine of one acute angle is equal to the cosine of the other acute angle. This is because $\sin(x) = \frac{\text{opposite}}{\text{hypotenuse}}$ and $\cos(y) = \frac{\text{adjacent}}{\text{hypotenuse}}$. Since the side opposite angle $x$ is the side adjacent to angle $y$, the two ratios are equal.

Step 2: Find the cosine of $y$

Given that $\sin(x) = 0.6$, we can directly find the cosine of $y$ using the relationship described in the previous step:

$\cos(y) = \sin(x)$

Therefore, $\cos(y) = 0.6$.

Step 3: Solve the cubic equation

The given equation is $x^3 - 5x^2 + 2x - 10 = 0$. We can try to factor this equation by grouping:

$x^2(x - 5) + 2(x - 5) = 0$

$(x^2 + 2)(x - 5) = 0$

This gives us two factors: $x^2 + 2 = 0$ and $x - 5 = 0$.

The first factor, $x^2 + 2 = 0$, gives $x^2 = -2$, which has no real solutions.

The second factor, $x - 5 = 0$, gives $x = 5$.