Question

Can someone please show me how to find all the solutions to this equation:
7 sin^2x- 14sinx+ 2= -5

Answer 1

Do this like you would a "normal" polynomial. In fact you could use a "u" substitution to make it easier. Let u = sin(x), so u^2 would equal sin^2(x). We have then a simple polynomial. \[7u ^{2}-14u+7=0\]If you factor out a 7 you get\[7(u ^{2}-2u+1)=0\]This would factor into \[7(u-1)(u-1)=0\]or \[7(u-1)^{2}=0\]Now fill the sin(x) back in and get \[\sin(x)-1=0\]Add 1 to both sides and get sin(x) = 1 (forget the 7 here because 7 will never equal 1!). Now the question is, where does sin(x) = 1? The unit circle will tell you that the sin(x) = 1 at pi/2 which is 90 degrees.

Answer 2

Ah, if it was just substituting the entire time I would've done that ages ago! Thank you :)
One more question: would you use that same method if instead of sinx it were to be cosx?

Answer 3

Yes the same method can be used for this problem type regardless of the trig function. If it has a form similar to a quadratic equation.

Answer 4

Ah thank you :)

Answer 5

You're welcome