Question

ayuda necesito la respuesta y el procedimiento porfa (secx-cosx)cosx=sen²x

Answer 1

( sec(x) - 1/sec(x) )( 1/sec(x) ) = 1/2 - (1/2)( 1/sec(x) )

( sec(x) - 1/sec(x) )( 1/sec(x) ) - 1/2 + (1/2)( 1/sec(x) ) = 0

(sec(x)² - 1)/(sec(x))(1/sec(x)) - 1/2 + 1/(2sec(x)) = 0

(sec(x)² - 1)/(sec(x)²) - 1/2 + 1/(2sec(x)) = 0

(2(sec(x)² - 1) - sec(x)² + sec(x))/(2sec(x)²) = 0

2(sec(x)² - 1) - sec(x)² + sec(x) = 0

sec(x)² - 2 + sec(x) = 0

t² - 2 + t = 0
t = 1 o t = - 2

sec(x) = 1
sec(x) = - 2

x = 2kπ
x = (2π/3) + 2kπ
x = (4π/3) + 2kπ

x = 2kπ/3, con k∈Z