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Solve for $x$, $\tan x +\sec x = 2\cos x$ ; $−∞ < x < ∞$

$$\tan x + \sec x = 2\cos x$$

I tried changing it all to sin and cos

$$\frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2\cos x$$

then I made it to one fraction

$$\frac{\sin x + 1}{\cos x} = 2 \cos x$$

Then I don't know where to go from there. Please help!

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  • $\begingroup$ Next cross multiply, treat where cos=0 separately. $\endgroup$ – coffeemath Jul 25 '14 at 3:27
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That gives you $$\sin x= 2\cos^2x-1 = \cos(2x).$$ It should be easy from here.

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  • $\begingroup$ thank you so much, I got it now $\endgroup$ – K. Wald Jul 25 '14 at 3:43
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$$(1+\sin x)=2\cos^2x=2(1+\sin x)(1-\sin x)$$

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$$\sec x+\tan x=2\cos x\iff \sec x-\tan x=\frac1{2\cos x}$$

Adding we get $$2\sec x=2\cos x+\frac1{2\cos x}$$

Multiplying by $2\cos x,$

$$4=4\cos^2x+1\iff2\cos^2x=\frac32\iff\cos2x=2\cos^2x-1=\frac12=\cos60^\circ$$

$$\iff 2x=360^\circ m\pm60^\circ$$

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Multiply across by $\cos x$ to get $\sin x + 1 = 2 \cos^2 x= 2 - 2 \sin^2 x$, or $2 \sin^2 x +\sin x -1 = 0$.

Since the solutions of $2y^2+y-1 = 0$ are $\{-1, {1 \over 2} \}$, the possible solutions to the original equations are $\sin x = -1$ and $\sin x = {1 \over 2} $. However, if $\sin x = -1$, then $\cos x = 0$ and the equation is undefined, hence we ignore this possibility.

The remaining solutions are $\arcsin ( { 1\over 2}) + 2\pi n= { \pi \over 6} + 2 \pi n$.

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