tan x + sec 2x = 1
asked Nov 23, 2016 in Trigonometry Answers by Ram

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1 Answer

When x=0: tan(0)+sec(2*0)=1 so x=0 is a solution, because tan(0)=0 and cos(0) and sec(0)=1.

There is another solution at x=67.5 degrees. See below:

If y=sin(x) then we can work out tan(x). cos(2x)=1-2sin^2(x)=1-2y^2, sec(2x)=1/(1-2sin^2(x)); cos(x)=√(1-y^2). sin=opp/hyp=y/1 so adj=√(1-y^2). tan(x)=opp/adj=y/√(1-y^2).

Therefore, tan(x)+sec(2x)=y/√(1-y^2)+1/(1-2y^2)=1.

y/√(1-y^2)=1-1/(1-2y^2)=(1-2y^2-1)/(1-2y^2)=-2y^2/(1-2y^2).

y=0 is a solution (sin(x)=0 so x=0) and:

1/√(1-y^2)=-2y/(1-2y^2); √(1-y^2)=(2y^2-1)/2y.

Squaring:

1-y^2=(4y^4-4y^2+1)/4y^2; 4y^2-4y^4=4y^4-4y^2+1; 8y^4-8y^2+1=0.

So y^2=(8±√(64-32)/16=1/2±√2/4=0.8536 or 0.1464 approx and y=0.9239 or 0.3827=sin(x).

Therefore x=±67.5º, x=±22.5º, but we may have to eliminate "solutions" that don't fit the original equation.

These are the solutions that do fit: 67.5, -22.5 and of course 0. The periodic nature of the trig functions gives us a series: 0, 67.5, -22.5, 157.5, 180, 247.5, 337.5, etc.

 

answered Nov 23, 2016 by Rod Top Rated User (486,900 points)
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