2tan2xcos2x+root3=0 for-180<=x<=180

Answer 1

To solve the equation 2tan(2x)cos(2x) + √3 = 0 for -180° ≤ x ≤ 180°, we can use the following steps:

First, let's multiply through by cos(2x) to simplify the equation:

2tan(2x)cos(2x)cos(2x) + √3cos(2x) = 0

Next, we can apply the trigonometric identity tan(2x) = sin(2x) / cos(2x):

2(sin(2x) / cos(2x))cos(2x)cos(2x) + √3cos(2x) = 0

Simplifying further, we have:

2sin(2x)cos(2x) + √3cos(2x) = 0

Now, we can factor out the common factor of cos(2x):

cos(2x)(2sin(2x) + √3) = 0

Using the zero product property, we set each factor equal to zero:

cos(2x) = 0 or 2sin(2x) + √3 = 0

For cos(2x) = 0, we have the solutions:

2x = 90° + 180°n or 2x = 270° + 180°n (where n is an integer)

Simplifying, we get:

x = 45° + 90°n or x = 135° + 90°n

For 2sin(2x) + √3 = 0, we can solve for sin(2x) as follows:

2sin(2x) = -√3
sin(2x) = -√3/2

Using the inverse sine function, we find the possible values for 2x:

2x = -60° + 360°n or 2x = 240° + 360°n (where n is an integer)

Simplifying, we get:

x = -30° + 180°n or x = 120° + 180°n

Therefore, the solutions to the equation 2tan(2x)cos(2x) + √3 = 0 for -180° ≤ x ≤ 180° are:

x = 45° + 90°n, 135° + 90°n, -30° + 180°n, 120° + 180°n