Verify that her solutions are correct.

Question

Latoya determined two approximate solutions to the equation 3 + sin² x = tan² x + 1 to be these: x = 1.027 and x = 2.281. Verify that her solutions are correct. Explain your reasoning. Can you also explain how to use sin² x on the calculator. Thank you.

Answer 1

Answer 2

The equation to solve is 3 + sin² x = tan² x + 1.

We know that tan(x) = sin(x)/cos(x), so the equation can be rewritten 3 + sin² x = (sin² x/cos² x) + 1.

The right-hand sided can be rewritten

(sin² x/cos² x) + (cos² x/cos² x)

On the previous line, we can factor out 1/cos² x from both terms, so the right-hand side of our equation becomes

(sin² x + cos² x)/cos² x = 1/cos² x.

But sin² x + cos² x = 1 implies sin² x = 1 - cos² x, so the left-hand side of our equation becomes

3 + sin² x = 3 + (1 - cos² x) = 4 - cos² x

Substituting in the new expressions for the left-hand and right-hand sides of our equation,

4 - cos² x = 1/cos² x

Let z = cos² x, then our equation becomes

4 - z = 1/z

Multiplying both sides by z

z(4 - z) = 1

Multiplying out the left-hand side,

4z - z² = 1

Gathering everything on the right-hand side of the equation,

0 = z² - 4z + 1

We can use the quadratic formula to solve for z:

az² + bz + 1 = z² - 4z + 1

So a = 1, b = -4, and c = 1, and

z = (-b±(b² - 4ac)^1/2)/2a

= (4 ± (4² - 4*1*1)^1/2)/2

= (4 ± (16 - 4)^1/2)/2

= (4 ± 12^1/2)/2

= (4 ± (4*3)^1/2)/2

= (4 ± (4^1/2)*(3^1/2)/2

= (4 ± (2)*(3^1/2)/2

= 2 ± 3^1/2

So we have two solutions for z:

1. 2 + 3^1/2 ≈ 3.73205
2. 2 - 3^1/2 ≈ 0.26795

But z = cos² x, and cos(x) is always between -1 and +1, so z = cos² x is always between 0 and +1.

But 2 + 3^1/2 > 1, so the solution z = 2 + 3^1/2 is not valid, so

cos²x = z = 2 - 3^1/2 ≈ 0.26795

Taking the square root of both sides,

cos(x) ≈ ±0.51764

A crude approximation is cos(x) = ±0.5, which we know how to solve:

Limiting ourselves to the interval 0 ≤ x ≤ π,

x = π/3 ≈ 3.14/3 = 1.05 solves cos(x) = +0.5, and

x = 2π/3 ≈ 2*3.14/3 = 2.09 solves cos(x) = -0.5.

cos(x) ≈ ±0.51764 gives us

x ≈ 1.0267 and x ≈ 2.1149, which are close to x ≈ 1.05 and x ≈ 2.09.

This is different from Latoya's solutions, x ≈ 1.027 and x ≈ 2.281.