Question

I have a trig test tomorrow, and I'm so stuck on a lot of these. Please help me out with this one, thanks!
6cosx +7tanx= secx

Answer 1

\[6 cosx +7 tanx= secx \]
\[6 \cos x+ 7 sinx/\cos x= 1/ \cos x\]

Answer 2

\[6 \cos^2x+7 sinx =1\]
we know
cos^2 x+ sin^2 x=1
so
cos^2 x = 1- sin^2 x
substituting this in given equation
6(1-sin2^x)+7 sin x=1
6 - 6 sin^2 x+7 sin x=1
-6 sin^2 x+ 7 sin x+5=0
6 sin^2 x- 7 sin x-5=0
we know the standard quadratic equation
ax^2+bx +c=0
x is given by quadratic formula
\[x=(-b \pm \sqrt (b^2-4ac))/2a\]
here
\[sinx= (7 \pm \sqrt( 49-4*6*(-5))/12\]
\[\sin x= (7 \pm \sqrt 169)/12\]
\[\sin x= (7 \pm 13)/12\]
sin x can't be 20/12 ( sin x<=1)
sin x=-6/12
sin x= -1/2
\[x= 2npi -\pi/6\]
\[x=( 2n+1) \pi + \pi/6\]
n is an integer

Answer 3

x= 7pi/6, 11pi/6, - pi/6 and so on

Answer 4

did you get it

Answer 5

I just don't understand how you got [6 \cos^2x+7 sinx =1\], the rest makes sense to me.

Answer 6

we have 6cosx+7tanx=secx
sec x can be written as 1/ cos x
and tan x can be written as sinx/ cos x
so 6cosx+7sinx/cosx=1/cosx
now multiply both sides by cos x
6 cos^2x+ 7 sinx=1

Answer 7

Ohhhhh ok, thank you so much :)

Answer 8

welcome liz:)