Question

Question #e5f13

Answer 1

Solve `cos x + sin x = sec x`

Ans: x = 0 , `pi/4`

Explanation:

`f(x) = cos x + sin x - 1/cos x = 0`
`f(x) = cos^2 x - sin x.cos x - 1 = 0` (cos x not zero)
Apply trig identity:`cos ^2 x = (cos 2x - 1)/2`
and sin 2x = 2sin x.cos x
`(cos 2x + 1)/2 + sin2x/2 = 1`
cos 2x + sin 2x + 1 = 2
`sin 2x + cos 2x = 1 [cos a + sin a = sqrt2(cos a + pi/4)]`
`sqrt2sin (2x + pi/4) = 1`
`sin (2x + pi/4) = 1/sqrt2`-->
`2x + pi/4 = pi/4` and `2x + pi/4 = (3pi)/4`

a. `2x + pi/4 = pi/4` --> 2x = 0 --> `x = 0`
b. `2x + pi/4 = 3pi/4`--> `2x = pi/2` --> `x = pi/4`

Check
a. x = 0 --> cos x = 1, sin x = 0, sec x = 1 --> 1 = 1 OK

b. x = pi/4 --> cos (pi/4) + sin (pi/4) = 2/cos (pi/4)
`sqrt2/2 + sqrt2/2 = (2sqrt2)/2`OK