Find all the angles between 0° and 90° which satisfy the equation

sec²Θcosec²Θ + 2cosec²Θ = 8

multiply by sin^2 cos^2 to get

1 + 2 cos^2 = 8 sin^2 cos^2
8cos^4 - 6cos^2 + 1 = 0
(4cos^2 - 1)(2 cos^2 - 1)

so,

cos^2 = 1/4 or 1/2
cos = 1/2 or -1/2 or 1/√2 or -1/√2
skip the negative values, since we want 1st quadrant angles only

Θ = 45° or 60°

To find all the angles between 0° and 90° that satisfy the equation sec²Θcosec²Θ + 2cosec²Θ = 8, we can start by simplifying the equation.

1. Recall the trigonometric identities:
sec²Θ = 1 + tan²Θ
cosec²Θ = 1 + cot²Θ

2. Substitute these identities into the equation:
(1 + tan²Θ)(1 + cot²Θ) + 2(1 + cot²Θ) = 8

3. Simplify the equation:
(1 + tan²Θ)(1 + cot²Θ) + 2(1 + cot²Θ) = 8
(1 + tan²Θ)(1 + cot²Θ) + 2cot²Θ + 2 = 8
(1 + tan²Θ)(1 + cot²Θ) + 2cot²Θ = 6

4. Expand the brackets:
(1 + tan²Θ + cot²Θ + tan²Θcot²Θ) + 2cot²Θ = 6
1 + tan²Θ + cot²Θ + tan²Θcot²Θ + 2cot²Θ = 6

5. Combine the terms:
1 + 3tan²Θ + 3cot²Θ + tan²Θcot²Θ = 6

6. Rearrange the terms:
tan²Θcot²Θ + 4tan²Θ + 4cot²Θ = 5

At this point, we notice that the equation is quadratic in nature. We can make a substitution to simplify it further.

Let x = tan²Θ. Then, we can rewrite the equation as:
x² + 4x + 4/x = 5

Now, we have a quadratic equation that we can solve.

7. Multiply both sides of the equation by x to get rid of the denominator:
x³ + 4x² + 4 = 5x

8. Rearrange the equation to have it equal to zero:
x³ + 4x² - 5x + 4 = 0

9. Solve the equation x³ + 4x² - 5x + 4 = 0 using numerical or algebraic methods. Once you find the value(s) of x, substitute them back into the equation x = tan²Θ.

10. Solve for Θ using the equation tan²Θ = x. Take the square root of x to find the positive and negative values of tanΘ. Then, use the inverse tangent (tan⁻¹) function to find the angle Θ.

11. Check that the angles you find are between 0° and 90°.

Following these steps will help you find all the angles between 0° and 90° that satisfy the given equation.