to solve a trigo. equation

**ling_c_0202** · January 2nd 2007, 06:40 AM

How can I solve the equation

tan 2x + sec 2x = cos x + sin x

for 0 deg. <= x <= 360 deg. ?

**Soroban** · January 2nd 2007, 11:27 AM

Hello, ling_c_0202!

Here's one way . . . I'll start it off for you . . .

Solve: . $\tan2x + \sec2x \:=\:\cos x + \sin x$ .for $0^o \leq x \leq 360^o$

Square both sides: . $(\tan2x + \sec2x)^2 \:=\:(\cos x + \sin x)^2$

. . $\tan^2\!2x + 2\tan2x\sec2x + \sec^2\!2x\:=\:\cos^2\!x + 2\cos x\sin x + \sin^2\!x$

. . $\frac{\sin^2\!2x}{\cos^2\!2x} + 2\!\cdot\!\frac{\sin x}{\cos x}\!\cdot\!\frac{1}{\cos x} + \frac{1}{\cos^2\!2x} \:=\:\sin^2\!x + \cos^2\!x + 2\sin x\cos x$

. . $\frac{\sin^2\!2x + 2\sin2x + 1}{\cos^2\!x} \:=\:1 + \sin2x\quad\Rightarrow\quad\sin^2\!2x + 2\sin2x + 1$ $\:=\:\cos^2\!2x(1 + \sin2x)$

. . $\sin^2\!2x + 2\sin 2x + 1 \:=\:(1 - \sin^2\!2x)(1 + \sin2x)$

. . which simplifies to: . $\sin^32x + 2\sin^22x + \sin2x \:=\:0$

Factor: . $\sin2x\left(\sin^22x + 2\sin 2x + 1\right) \:=\:0\quad\Rightarrow\quad \sin2x(\sin2x + 1)^2\:=\:0$

And we have:
. . $\sin2x\:=\:0\quad\Rightarrow\quad 2x\:=\:0,\:\pi,\:2\pi,\:3\pi\quad\Rightarrow\quad\ boxed{x \:=\:0,\:\frac{\pi}{2},\;\pi,\:\frac{3\pi}{2},\:2\ pi}$

. . $\sin2x + 1 \:=\:0\quad\Rightarrow\quad\sin2x \:=\:-1\quad\Rightarrow\quad 2x \:=\:\frac{3\pi}{2},\:\frac{7\pi}{2}\quad\Rightarr ow\quad\boxed{x \:=\:\frac{3\pi}{4},\:\frac{7\pi}{4}}$

I'll let you check for extraneous roots . . .

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