(2sinx)+1/(2cosx)-√3=0

Предыдущий
вопрос Следующий
вопрос

To solve the equation, we need to simplify and rearrange the terms:

2sinx + 1/(2cosx) - √3 = 0

First, let's get a common denominator for sinx and cosx:

2sinx + 1/(2cosx) - √3 = 0
2sin^2x/2 + 1/(2cosx) - √3 = 0

Now, rewrite sin^2x as (1 - cos^2x) using the Pythagorean identity sin^2x + cos^2x = 1:

(2 - 2cos^2x)/2 + 1/(2cosx) - √3 = 0
(2/cos^2x - 2) + 1/(2cosx) - √3 = 0

Now multiply through by 2cos^2x to get rid of the fractions:

2cos^2x(2/cos^2x - 2) + 2cos^2x(1/(2cosx)) - 2cos^2x√3 = 0
4 - 4cos^2x + cosx - 2√3cos^2x = 0
4 - 6cos^2x + cosx = 0

Now, rearrange the terms to get a quadratic equation:

6cos^2x - cosx - 4 = 0

Now you can solve this quadratic equation for cosx using the quadratic formula or factoring.