trigonometric equations of quadratic type

[awstuck]

find all real number in the interval [0,2?) that satisy each equation. Round approximate answers to the nearest tenth.

cos(2X)cos(X)-sin(2X)sin(X)=1/2

1.) cos(2X+x)=1/2
cos(3X)=1/2
3X=cos^-1(1/2)
X=(cos^-1(1/2))/3
X=(5?/3)/3 X=(?/3)/3
X= 5?/9 X=?/9

OR
2.) cos(2X)cos(X)-2sin(x)cos(x)sin(x)=1/2
cos(x)(cos(2x)-2sin(x)sin(x))=1/2
cos(x)=1/2 ; cos(2x)-2sin^2(x)=1/2
x=?/3 x=5?/3 ; 1-2sin^2(x)=1/2
-2sin^2(x)-2sin^2(x)=-1/2
-2(sin^2(x)-sin^2(x))=1/2
sin^2(x)+sin^2(x)=1/4
2sin^2(x)=1/4
sin^2(x)=1/8
sin(x)=sqrt(1/8) sin(x)=-sqrt(1/8)
sin(x)=1/(2sqrt(2)) sin(x)=-1/(2sqrt(2))
x=sin^-1(1/(2sqrt(2))) x=sin^-1(-1/(2sqrt(2)))
OR

3.) (cos^2(x)-sin^2(x))cos(x)-(2sin(x)cos(x))sin(x)=1/2
cos^3(x)-sin^2(x)cos(x)-2sin^2(x)cos(x)=1/2
cos(x)(cos^2(x)-sin^2(x)-2sin^2(x))=1/2
cos(x)=1/2 cos^2(x)-3sin^2(x)=1/2
x=?/3 x=5?/3

OR

4.)cos(2x)cos(x)-sin(2x)sin(x)=1/2
(2cos^2(x)-1)cos(x)-(2sin(x)cos(x))sin(x)=1/2
2sin^3(x)-cos(x)-2sin^2(x)cos(x)=1/2
2cos^3(x)-cos(x)-2(1-cos(2x)/2)cos(x)=1/2
2cos^3(x)-cos(x)-(1-cos(2x)cos(x))=1/2
2cos^3(x)-cos(x)-2cos^2(x)cos(x)=1/2
-cos(x)=1/2
cos(x)=-1/2
x=2?/3 x=4?/3

 

[galactus]

find all real number in the interval [0,2?) that satisy each equation. Round approximate answers to the nearest tenth.

\(\displaystyle cos(2X)cos(X)-sin(2X)sin(X)=\frac{1}{2}\)

X= 5?/9, X=?/9

In the interval \(\displaystyle [0, \;\ 2\pi]\), there are 6 solutions that result in 1/2:

\(\displaystyle \frac{\pi}{9}, \;\ \frac{5\pi}{9}, \;\ \frac{7\pi}{9}, \;\ \frac{11\pi}{9}, \;\ \frac{13\pi}{9}, \;\ \frac{17\pi}{9}\)

 

[soroban]

Hello, awstuck!

I agree with galactus.

\(\displaystyle \text{Find all real number in the interval }[0,\,2\pi)\text{ that satisy each equation.}\)

. . . \(\displaystyle \cos2x\cos x - \sin2x\sin x \:=\: \tfrac{1}{2}\)

\(\displaystyle \text{We have the identity: }\:\cos A\cos B - \sin A\sin B \:=\:\cos(A+B)\)

\(\displaystyle \text{The equation becomes: }\;\cos3x \:=\:\tfrac{1}{2}\)
. . \(\displaystyle \text{Hence: }\;3x \:=\:\pm\frac{\pi}{3} + 2\pi n \quad\Rightarrow\quad x \:=\:\pm\frac{\pi}{9} + \frac{2\pi}{3}n\)


\(\displaystyle \text{For }n = 0\!:\;\;x \;=\qquad\pm\frac{\pi}{9}\quad \;=\; \left\{\frac{\pi}{9}\right\}\)

\(\displaystyle \text{For }n = 1\!:\;\;x \;=\;\pm\frac{\pi}{9} + \frac{2\pi}{3} \;=\;\begin{Bmatrix}\dfrac{5\pi}{9} \\ \\[-2mm] \dfrac{7\pi}{9}\end{Bmatrix}\)

\(\displaystyle \text{For }n = 2\!:\;\;x \;=\;\pm\frac{\pi}{9} + \frac{4\pi}{3} \;=\;\begin{Bmatrix}\dfrac{11\pi}{9} \\ \\[-2mm] \dfrac{13\pi}{9}\end{Bmatrix}\)

\(\displaystyle \text{For }n = 3\!:\;\;x \;=\;\pm\frac{\pi}{9} + 2\pi} \;=\; \left\{\frac{17\pi}{9}\right\}\)

 

[BigGlenntheHeavy]

\(\displaystyle Good \ show \ soroban, \ the \ identity \ was \ the \ clincher.\)