i'm lost...how would i solve for x given: 6cosx^2 + sinx - 4=0 ???

Question

is there some trick to it that i just have yet to learn??
p.s. i know i ask a lot of math questions...THANKS TO THOSE OF YOU WHO CONTINUE TO HELP ME!

Answer 1

Have you learned trig identities yet? I'd check to see if there was a trig identity you could substitute here.

BTW, I am assuming you meant cos²x rather than cosx².

Try this: cos²x + sin²x = 1, so substitute 1-sin²x for cos²x:

6cos²x + sinx - 4 = 0

6(1-sin²x) + sinx - 4 = 0

-sin²x + sinx + 2 = 0

(multiply by -1 to make it a little easier)

sin²x - sinx - 2 = 0

(sinx-2)(sinx+1) = 0

So, sinx = 2 or sinx = -1

Solving for x, this means x = sin^-1(2) or sin^-1(-1)

Answer 2

Substitute 1-sin²(x) for cos²(x), multiply through by the 6 to get 6 -6sin²(x) + sin(x) - 4 = 0 Rewrite this as
6sin²(x) -sin(x) -2 = 0 and solve the quadratic for sin(x). Then x is arcsin of whatever you got for sin(x)


Doug

Answer 3

cos^2x = 1 - sin^2x. then you'll get a quadratic equation, which can either be factored or solved with the quadratic formula. then you get sinx = -2/3 and 1/2. then x = sin^-1 of those 2 values.

Answer 4

cos^2(x)+sin^2(x)=1, use this to find a substitution for cosine squared, and your equation will become a standard quadratic equation.