Question

1)4cos^2x=3 2)4sin^2x-cosx-1=0

Answer 1

1)4cos²x = 3

2cos²x = 3/2

1+cos2x = 3/2

cos2x = 1/2

2x = ±π/3 + 2πk

x = ±π/6 + πk

2)4sin²x - cosx - 1 = 0

4(1-cos²x) - cos x - 1 = 0

4-4cos²x - cosx - 1 = 0

4cos²x + cosx - 3 = 0

cos x = a

4a²+a-3= 0

a1 = -1

a2 = 3/4

a)cos x = -1

x = π + 2πk

б)cos x = 3/4

x = ±arccos(3/4) + 2πk

Answer 2

Let's solve each equation one by one:

1. 4cos^2(x) = 3

To solve this equation, we'll manipulate it using trigonometric identities. We know that cos^2(x) + sin^2(x) = 1, so we can substitute 1 - sin^2(x) for cos^2(x):

4(1 - sin^2(x)) = 3

Distribute the 4:

4 - 4sin^2(x) = 3

Rearrange the equation:

4sin^2(x) = 4 - 3

4sin^2(x) = 1

Divide both sides by 4:

sin^2(x) = 1/4

Taking the square root of both sides:

sin(x) = ±√(1/4)

sin(x) = ±1/2

Now, we need to find the values of x where sin(x) equals ±1/2. These values can be found on the unit circle. The solutions are:

x = π/6 + 2πn, x = 5π/6 + 2πn (for sin(x) = 1/2) x = 7π/6 + 2πn, x = 11π/6 + 2πn (for sin(x) = -1/2)

where n is an integer.

2. 4sin^2(x) - cos(x) - 1 = 0

This equation is a bit more complicated. Let's solve it step by step:

Rearrange the equation:

4sin^2(x) - cos(x) = 1

Apply the trigonometric identity sin^2(x) + cos^2(x) = 1:

4(1 - cos^2(x)) - cos(x) = 1

Distribute the 4:

4 - 4cos^2(x) - cos(x) = 1

Rearrange the equation:

4cos^2(x) + cos(x) - 3 = 0

Now we have a quadratic equation in terms of cos(x). We can solve it using factoring or the quadratic formula. Let's use factoring:

(4cos(x) - 3)(cos(x) + 1) = 0

Setting each factor equal to zero:

4cos(x) - 3 = 0 or cos(x) + 1 = 0

Solving the first equation:

4cos(x) = 3

cos(x) = 3/4

The second equation:

cos(x) = -1

Now, we need to find the values of x where cos(x) equals 3/4 or -1. The solutions are:

x = arccos(3/4) + 2πn, x = -arccos(3/4) + 2πn (for cos(x) = 3/4) x = π + 2πn (for cos(x) = -1)

where n is an integer.