solve tan^2x+secx =1 in the range 0°≤x≤ 360°

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

tan^2 x +sec x=1

We know that tan=sin/cos  and sec= 1/cos

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

we know that: sin^2 = 1-cos^2

==> [(1-cos^2 x)/cos^2 x]+ 1/cos x = 1

==> 1-cos^2 x +cos x = cos^2 x

==> 2cos^2 x -cos x -1 =0

==> (2cosx +1)(cos x -1)=0

==> cosx=1 ==> x= 0 , 360

==> 2cosx+1=0 ==> cosx= -1/2 ==> x= 120 degrees, 240 degrees

==> x = {0, 120, 240, 360}

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

sec x = 1/cos x and (tan x)^2 = (sinx/cosx)^2

From the fundamental formula of trigonometry, (sin x)^2 = 1-(cos x)^2.

(sinx/cosx)^2 = (1-(cos x)^2)/(cos x)^2

The expression will become:

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

We'll do the math so that all terms of the both sides of expression to have the same denominator, which is (cos x)^2.

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

We'll move all terms to one side:

1-(cos x)^2 + cos x - (cos x)^2 = 0

We'll add the similar terms:

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

We'll note cos x = t

-2t^2 + t + 1 = 0

We'll multiply by (-1):

2t^2 - t - 1 = 0

Being a quadratic equation, we'll use the formula:

x = [-b+sqrt(b^2-4ac)]/2a, where a=2, b=-1 and c=-1.

t1 = [1+sqrt(1+4*2)]/2*2

t1 = (1+3)/4

t1 = 1

t2 = (1-3)/4

t2 = -2/4

t2 = -1/2

Let's recall that cos x = t

cos x = t1

cos x = 1

It's an elementary equation:

x = 0 or x = 360, in the interval [0,360]

cos x = -1/2

The function cosine is negative in the quadrant 2 and 3, so

x = 90+60 = 150

x = 180 + 60 = 240

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