Im stuck on this trig problem can anyone help?

Question

tanx + secx = 2cosx

Answer 1

tanx + secx = 2cosx

sinx/cosx + 1/cosx = 2cosx

(sinx +1)/cosx = 2cosx

sinx+1 = 2cos^2x

2(1-sin^2x) = sinx +1

2-2sin^2x = sinx +1

0 = 2sin^2x+sinx-1

Now it's a quadratic equation:

a=2 b=1 and c=-1 and place into quadratic equation such that

sinx=(1/2) and -1

or x = invsin(1/2) and x = invsin (-1)

Answer 2

tan(x) + sec(x) = 2cos(x)

sin(x)/cos(x) + 1/cos(x) = 2cos(x)

sin(x) + 1 = 2cos²(x) ------ cos(x)≠0

sin(x) + 1 = 2 - 2sin²(x)

2sin²(x) + sin(x) - 1 = 0

2sin²(x) - sin(x) + 2sin(x) - 1 = 0

sin(x)(2sin(x)-1) + 1(2sin(x)-1) = 0

(sin(x)+1)(2sin(x)-1) = 0

sin(x)+1 = 0

sin(x) = -1

This means that cos(x) = 0

This will not lead to a valid solution

2sin(x) - 1 = 0

sin(x) = 1/2

x = {2nπ + π/6, 2nπ + 5π/6} where n is an integer

Answer 3

tan x+sec x = 2cos x -----------------------------the problem

sin x/cos x+sec x = 2cos x --------------------tan x = sin x / cos x

−2cos x+sec x+ sin x/cos x= 0 --------------subtract 2 cos x

−2cos x+ 1/cos x+ sin x/cos x= 0 ----------sec x = 1 / cos x

−2cos x+1+sin x/cos x= 0 ---------------------a/c + b/c = (a+b) /c

−2cos x+sin x+1/cos x= 0 ---------------------put terms in order

sin x−2cos^2 x+1/cos x= 0 --------------------common denom and simp

sin x−2cos^2 x+1 = 0 ----------------------------multiply by cos x

sin x−2(1−sin^2 x ) +1 = 0 ---------------------cos^2 u = 1 - sin^2 u

u−2(1−u^2)+1 = 0 ----------------------------------u = sin x

2u^2 +u−1 = 0 ---------------------------------------s...

(u+1)(2u−1) = 0 -------------------------------------fac... quadratic

u = −1 ------------ u =1/2 ---------------------------a=0 or b=0 if ab=0

sin x = -1 --------- sin x =1/2 --------------------u= sin x