## cupcakezz one year ago 2cos^2x + sinx-1=0 a. Explain why this equation cannot be factored. b. Use a trigonometric identity to change the equation into one that can be factored. c. Factor the equation. d. Determine all solutions in the interval 0≤x≤2π.

1. cupcakezz

@Luigi0210

2. RaphaelFilgueiras

|dw:1389329041541:dw|

3. cupcakezz

@bagajr ??

4. cupcakezz

can you send some things so I can start looking at it ?? @bagajr

5. cupcakezz

@kc_kennylau

6. kc_kennylau

Firstly, be familiarized with this identity: $$\sin^2x+\cos^2x=1$$

7. kc_kennylau

Corollary: $$\sin^2x=1-\cos^2x$$ and $$\cos^2x=1-\sin^2x$$.

8. kc_kennylau

You may want to convert the question into an ordinary quadratic equation.

9. cupcakezz

But why cant it be factored??

10. cupcakezz

@bagajr

11. kc_kennylau

You will know why when you convert the question into an ordinary quadratic equation by changing $$\cos^2x$$ to $$1-\sin^2x$$.

12. bagajr

a) Because cos^2x and sinx are not the same variable. b) Use the following identity to solve for cos^2x$\sin^{2}(x)+\cos^{2}(x)=1$$\cos^{2}(x) =1-\sin^{2}(x)$Substitute cos^2x for this new expression.$2\cos^{2}(x)+\sin(x)-1=0$$2[1-\sin^{2}(x)]+\sin(x)-1=0$$2-2\sin^{2}(x)+\sin(x)-1=0$$-2\sin^{2}(x)+\sin(x)+1=0$$2\sin^{2}(x)-\sin(x)-1=0$c) This isn't necessary, but you can make factoring easier by setting y=sin(x)$2y^{2}-y-1=0$$(2y+1)(y-1)=0$$(2\sin(x)+1)(\sin(x)-1)=0$d) When you multiply two things to get 0, one of those things must be 0.$2\sin(x)+1=0$$\sin(x)-1=0$Solve both equations to get that sin(x) is -1/2 or 1. Use unit circle to find solutions, which are pi/2, 5pi/4, and 7pi/4. |dw:1389329795828:dw|

13. cupcakezz

@bagajr wow thank you very much !

14. bagajr

No problem, I love helping people with this stuff.

15. cupcakezz

@bagajr could you help me with one more?

16. bagajr

Is it a question about this problem or a different one?

17. cupcakezz

prove: tan^2x(1+1/tan^2x)=1/(1-sin^2x)

18. cupcakezz

@bagajr

19. cupcakezz

@bagajr what is the answer??

20. bagajr

$\tan^2(x)[1+\frac{1}{\tan^{2}(x)}]=\frac{1}{1-\sin^{2}(x)}$$\tan^{2}(x)+1=\frac{1}{\cos^{2}(x)}$$\frac{\sin^{2}(x)}{\cos^{2}(x)}+1=\frac{1}{\cos^{2}(x)}$$\sin^{2}(x)+\cos^{2}(x)=1$$1=1$

21. cupcakezz

thanks again !

22. bagajr

Sorry it takes so long and you're welcome

23. cupcakezz

its fine, no worries