a) 2 cos?x - 3 cosx + 1 = 0; 6) 2 sin°x + sinx -1=0

Let's solve each equation step-by-step.

a) \( 2 \cos^2 x - 3 \cos x + 1 = 0 \)

This is a quadratic equation in terms of \( \cos x \). Let \( u = \cos x \). Thus, the equation becomes:

\[ 2u^2 - 3u + 1 = 0 \]

To solve this, we can use the quadratic formula:

\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Where \( a = 2 \), \( b = -3 \), and \( c = 1 \).

Calculating the discriminant:

\[ b^2 - 4ac = (-3)^2 - 4 \cdot 2 \cdot 1 = 9 - 8 = 1 \]

Now, using the quadratic formula:

\[ u = \frac{3 \pm \sqrt{1}}{2 \cdot 2} = \frac{3 \pm 1}{4} \]

Calculating the two possible values:

1. \( u = \frac{3 + 1}{4} = 1 \)
2. \( u = \frac{3 - 1}{4} = \frac{1}{2} \)

Now revert back to \( \cos x \):

1. \( \cos x = 1 \) \(\implies x = 0 + 2k\pi \) (where \( k \) is any integer)
2. \( \cos x = \frac{1}{2} \) \(\implies x = \frac{\pi}{3} + 2k\pi \text{ or } x = \frac{5\pi}{3} + 2k\pi \) (where \( k \) is any integer)

So, the solutions for \( x \) in this case are:

• \( x = 0 + 2k\pi \)
• \( x = \frac{\pi}{3} + 2k\pi \)
• \( x = \frac{5\pi}{3} + 2k\pi \)

b) \( 2 \sin^2 x + \sin x - 1 = 0 \)

This is also a quadratic equation, but in terms of \( \sin x \). Let \( v = \sin x \). Thus, we get:

\[ 2v^2 + v - 1 = 0 \]

Using the quadratic formula again:

\[ v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Where \( a = 2 \), \( b = 1 \), and \( c = -1 \).

Calculating the discriminant:

\[ b^2 - 4ac = 1^2 - 4 \cdot 2 \cdot -1 = 1 + 8 = 9 \]

Now substituting into the quadratic formula:

\[ v = \frac{-1 \pm \sqrt{9}}{2 \cdot 2} = \frac{-1 \pm 3}{4} \]

Calculating the two possible values:

1. \( v = \frac{2}{4} = \frac{1}{2} \)
2. \( v = \frac{-4}{4} = -1 \)

Now revert back to \( \sin x \):

1. \( \sin x = \frac{1}{2} \) \(\implies x = \frac{\pi}{6} + 2k\pi \text{ or } x = \frac{5\pi}{6} + 2k\pi \) (where \( k \) is any integer)
2. \( \sin x = -1 \) \(\implies x = \frac{3\pi}{2} + 2k\pi \) (where \( k \) is any integer)

So, the solutions for \( x \) in this case are:

• \( x = \frac{\pi}{6} + 2k\pi \)
• \( x = \frac{5\pi}{6} + 2k\pi \)
• \( x = \frac{3\pi}{2} + 2k\pi \)

Summary of Solutions

For equation a:

• \( x = 0 + 2k\pi \)
• \( x = \frac{\pi}{3} + 2k\pi \)
• \( x = \frac{5\pi}{3} + 2k\pi \)

For equation b:

• \( x = \frac{\pi}{6} + 2k\pi \)
• \( x = \frac{5\pi}{6} + 2k\pi \)
• \( x = \frac{3\pi}{2} + 2k\pi \)

Feel free to ask if you need further clarification or assistance!