Let's split the given equation (2cosx-1)(2sinx+√3 ) = 0 into two parts, and solve each separately. These parts would be 2cos(x) - 1 = 0, and 2sin(x) + √3 = 0.

$2\cos \left(x\right)-1=0,\\2\cos \left(x\right)=1,\\\cos \left(x\right)=\frac{1}{2}$

Remember that the general solutions for cos(x) = 1/2 are x = π/3 + 2πn and x = 5π/3 + 2πn. In this case we are given the interval 0 ≤ x ≤2π, and therefore x = π/3, and x = 5π/3.

Similarly:

$\:2\sin \left(x\right)+\sqrt{3}=0,\\2\sin \left(x\right)=-\sqrt{3},\\\sin \left(x\right)=-\frac{\sqrt{3}}{2}$

The general solutions for sin(x) = - √3/2 are x = 4π/3 + 2πn and x = 5π/3 + 2πn. Therefore x = 4π/3 and x = 5π/3 in this case.

So we have x = π/3, x = 5π/3, and x = 4π/3 as our solutions.

5 0

3 years ago

PLZ HELP its Triangle Properties and where using the 45º-45º-90º Triangle Theorem and the 30º-60º-90º Triangle Theorem.

den301095 [7]

Answer:

The answer to your questions is below

Step-by-step explanation:

4.-

In a 45°-45°-90° triangle, the length of the legs are "x" and the length of the hypotenuse is x $\sqrt{2}$

Then, in this problem, the length of the leg is 7 then the length of the hypotenuse will be 7 $\sqrt{2}$

5.- In a 30°-60°-90° triangle, the length of the hypotenuse is 2x, the length of the longer leg is x $\sqrt{3}$ and the length of the shorter leg is x.

In this problem we have the length of the hypotenuse = 20 and find x.

2x = 20

x = 20/2

x = 10

longer leg = x $x\sqrt{3} = 10\sqrt{3}$

6.- nonzero

7.- hypotenuse = 20 cm

Find x

2x = 20

x = 20/2

x = 10

Shoter leg = x = 10 cm

4 0

3 years ago

Read 2 more answers