please help me solve this trig equation?

Question

solve: tan^2x + secx = 1

I know tan x = sinx/cosx does this mean tan^2x would be (sinx/cosx)^2 ??

and secx = 1/cosx

Please help! (: im confused with the trig identities part but im good with the quadrant and finding angles :)

Answer 1

THE OTHER ANSWERS HERE ARE INCORRECT!

You are absolutely correct: tan^2(x) = (sinx/cosx)^2, because you are simply squaring both sides of that identity.

Anyway, on with the question:

tan^2(x) + sec(x) = 1

whenever you seen weird things, its probably best to write it in terms of sin and cos and see what happens.

(sinx/cosx)^2 + (1/cosx) = 1

lets multiply both sides by cos^2(x)

sin^2(x) + cosx = cos^2(x)

okay so at the moment we have only 2 different terms: sin and cos. lets try to get rid of that sin. recall that:
sin^2(x) + cos^2(x) = 1.
so:
sin^2(x) = 1- cos^2(x).
let's substitute this into our equation.

1- cos^2(x) + cosx = cos^2(x)

2cos^2(x) - cosx - 1 = 0

this is a quadratic equation, so we can solve for cosx

you can either use the quadratic formula or factorise, but you should come up with:

cosx = -(1/2) and cosx = 1

So from here you can use the quadrants to solve for x.

So just for the interval between 0 and 360 degrees:

x = 0, 120, 240, 360

Hope this helps!

Answer 2

Put 1 - cos^2 in place of sin^2, then let m = cos x. This will leave you with the quadratic equation m^2 + 4m - 1 = 0. Solve for m then put cos(x) back in place and solve for x.

Answer 3

Hello

Also note that tan^2(x) = sec^2(x) -1

So substitute this in to get

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

Simplified to be

sec^2(x) + sec(x) - 2 = 0
This is simply

(u+2)(u-1) = 0 (let u = secx)

Thus we have

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

So we have
(sec(x)+2) = 0
or
(sec(x)-1) = 0

sec(x) = 2 when x = pi/3 and 5pi/3
sec(x) = 1 when x = 0 and 2pi

(this is also the same thing as finding where cos(x) = 1 and cos(x)=.5

Hope this helps

Answer 4

One of the trig identities is
1 + tan^2 x = sec^2 x
so...
tan^2 x = (sec^2 x) - 1

So, use that substitution in the equation...

tan^2 x + sec x = 1
(sec^2 x - 1) - sec x = 1
sec^2 x - sec x - 2 = 0

Let y = sec x
now you have

y^2 - y - 2 = 0
(y - 2)(y + 1) = 0
so,
y = 2 and -1
which means

sec x = 2 and sec x = -1
so...
x = 60 and x = 180

Answer 5

yes, tan^2 x = (sinx / cosx) ^ 2

Let sinx = a, cosx = b

we have a^2 / b^2 + 1/b = 1
or a^2 + b = b^2

note that a^2 = 1-b^2

so 1 - b^2 + b = b^2
or 2b^2 -b-1 = 0
(2b+1)(b-1) = 0

so we have b = 1 and x = 0
or b = -1/2 and x = 2pi/3 or -2pi/3

Answer 6

Yes, tan^2 x = (sin x / cos x)^2 or sin^2/cos^2

and

sec x = 1/cos x

So...

tax^2 + sec =1
(sin^2/cos^2) + (1/cos) = 1
Getting common denominators leaves:
(sin^2/cos^2) + (cos/cos^2) = 1
Adding the fractions leaves:
(sin^2 + cos ) / cos^2 = 1
Therefore:
sin^2 + cos = cos^2

Since sin^2 + cos^2 =1, sin^2 = 1 - cos^2

Back to our equation: sin^2 + cos = cos^2, substituting for sin^2 leaves:
1 - cos^2 + cos = cos^2.

Now, lets let m=cos x

1 - m^2 + m = m^2 or...
2m^2 - m - 1 = 0

so...
m = cos x = 1 or -0.5

if cos x = 1, x = 0

if cos x = -0.5, x = 120 degrees or (2 pi / 3) radians

x = 0 or 2pi/3

Answer 7

you are rught:
tan^2(x) = (sin x/cos x)^2 = sin^2x / cos^2x
& sec x = 1/cos x,
so it comes:
sin^2x / cos^2x + 1/cos x = 1 <=>
sin^2x / cos^2x + cos x/cos^2 x = 1 <=>
sin^2x + cos x = cos^2 x <=>
1-cos^2 x + cos x = cos^2 x
in the canonic form:
-cos^2 x + cos x + 1 = 0
consider cos x as y and solve:
-y^2 + y + 1= 0
and once you get the values for y, solve:
y = cos x

won't help you more than this

Answer 8

sin^2x/cos^2x+1/cos x-1=0
sin^2x +cos x-cos^2x=0 as sin^2 = 1-cos ^2
-2 cos^2x+cos x +1 =0
call cos x = z
2z^2-z-1=0 z=( 1+- 3)/4
so cos x= 1 and cos x = -1/2
x= 2kpi k any integer
x= 2pi/3+2kpi
x= 4pi/3 +2k pi

Answer 9

tanx=sinx/cosx
tan2x=sin2x/cos2x
sin2x=2cosxsinx
cos2x=1-2*(sinX)^2
------------------------------------------------------------------
tan^2x+secx=sin^2x/cos^2x+1/cos^2x=(1+sin^2x)/cos^2x=1
(because 1+sin^2x=cos^2x