Let f (x) = m vector × n vector, where vector M = (2cosx, 1), n vector = (cosx, √ 3sin2x), X belongs to R In the triangle ABC, a, B, C are the opposite sides of angles a, B and C respectively. We know that f (a) = 2, B = 1, and the area of triangle ABC is √ 3 / 2. We can find the value of B + C / SINB + sinc

Let f (x) = m vector × n vector, where vector M = (2cosx, 1), n vector = (cosx, √ 3sin2x), X belongs to R In the triangle ABC, a, B, C are the opposite sides of angles a, B and C respectively. We know that f (a) = 2, B = 1, and the area of triangle ABC is √ 3 / 2. We can find the value of B + C / SINB + sinc

F (x) = m vector × n vector = (2cosx, 1) * (cosx, √ 3sin2x) = 2 (COX) ^ 2 + √ 3sin2x = 1 + cos2x + √ 3sin2x
= 1 + 2Sin (2x + 30 degrees)
From F (a) = 2, there is 1 + 2Sin (2a + 30 degrees) = 2, a = 60 degrees
The area of the triangle ABC is √ 3 / 2, 1 / 2 * bcsina = 1 / 2 * 1 * csin, 60 degrees = √ 3 / 2, C = 2
According to the cosine theorem, a ^ 2 = B ^ 2 + C ^ 2-2bccosa = 1 + 4-2 * 1 * 2 * 1 / 2 = 3, a = √ 3, C = 90 degrees, B = 30 degrees
B + C / SINB + sinc = 1 + 2 / sin30 degree + sin90 degree = 1 + 4 + 1 = 6