MHB Finding Points of Intersection for Trigonometric Functions

[osafi52]

does anybody know how to solve this problem ?Consider two trigonometric functions y = cos 2x and y = 1 + sinx.(a) Write down the equation in x that you would solve to find the x coordinate of point(s) of intersection of those graphs on [0, 2∏].

(b) Solve your equation, and write down the coordinates of the point(s) of intersection .

 

[Greg]

Re: need HELP on this question

Consider two trigonometric functions y = cos 2x and y = 1 + sinx.(a) Write down the equation in x that you would solve to find the x coordinate of point(s) of intersection of those graphs on [0, 2∏].

(b) Solve your equation, and write down the coordinates of the point(s) of intersection .

(a) $\cos(2x)=1+\sin(x)$

(b) $\cos(2x)=1-2\sin^2(x)=1+\sin(x)$
$2\sin^2(x)+\sin(x)=0$
$\sin(x)(2\sin(x)+1)=0$

Can you finish?

 

[osafi52]

thanks so much for the reply it really helped.once i solved the equation i got x=0 and x= 30 .

 

[skeeter]

thanks so much for the reply it really helped.once i solved the equation i got x=0 and x= 30 .

$$\sin(x) = 0$$ at $$x = 0^\circ$$ and $$x = 180^\circ$$ for $0^\circ \le x < 360^\circ$

$$2\sin(x)+1 = 0 \implies \sin(x) = -\dfrac{1}{2}$$ ... you want to try to find the degree solutions for $x$ again?

 

[osafi52]

Re: need HELP on this question

(a) $\cos(2x)=1+\sin(x)$

(b) $\cos(2x)=1-2\sin^2(x)=1+\sin(x)$
$2\sin^2(x)+\sin(x)=0$
$\sin(x)(2\sin(x)+1)=0$

Can you finish?

Hi. thanks for the reply. i solved the equation and got x =0 and x=-30. firstly is this correct and does this answer the question b.

 

[osafi52]

$$\sin(x) = 0$$ at $$x = 0^\circ$$ and $$x = 180^\circ$$ for $0^\circ \le x < 360^\circ$

$$2\sin(x)+1 = 0 \implies \sin(x) = -\dfrac{1}{2}$$ ... you want to try to find the degree solutions for $x$ again?

thanks for that. 2sinx+1=0 should have got x=-30.

 

[skeeter]

thanks for that. 2sinx+1=0 should have got x=-30.

how about x = 210 (or -150) ... ?

 

[tkhunny]

Just a thought question: Do "equations" intersect?

 

[Greg]

$$x\in\left\{0,\pi,\frac{7\pi}{6},\frac{11\pi}{6},2\pi\right\}$$