Question

Please solve all these I need urgent​

Answer 1

Answer:

2 cos x + 2√2 = 3 sec x

=> 2 cos x + 2√2 = 3/cos x

=> cos x ( 2 cos x + 2√2 ) = 3

=> 2 cos²x + 2√2 cos x = 3

=> 2 cos²x + 2√2 cos x - 3 = 0

put cos x = a

=> 2 a² + 2√2 a - 3 = 0

=> 2 a² + 3√2a - √2a - 3 = 0

=> √2*√2 a² + 3√2 a - √2 a - 3 = 0

=> √2a ( √2a + 3 ) - 1 ( √2a+ 3 ) = 0

=> (√2a - 1)(√2a + 3 ) = 0

=> √2 a - 1 = 0 and √2a + 3 = 0

therefore a = 1/√2. and a = - 3/√2

hence cos x = 1/√2 => cos x = cos 45° => x = 45° => Answer

a = - 3/√2 => cos x = -3/√2 × is rejected as x lies between 0° and 90° where cos x is positive