Precalc: Solve 2sinx^(2 ) - 5cosx - 4 = 0, 2cosx^(2) = 3sinx + 3

[Life_57]

The first problem I'm trying to solve for x and I only was able to get 2pi, 1pi/6, and 11pi/6 , but it says I'm wrong.

1. 2sinx^(2 )-5cosx-4=0

Problem 2 is similar, but still I don't know what I'm doing wrong. I used the Quadratic Formula and the Trig Identities

2. 2cosx^(2)=3sinx+3

 

[tkhunny]

Please demonstrate your work.

 

[Dr.Peterson]

The first problem I'm trying to solve for x and I only was able to get 2pi, 1pi/6, and 11pi/6 , but it says I'm wrong.

1. 2sinx^(2 )-5cosx-4=0

Problem 2 is similar, but still I don't know what I'm doing wrong. I used the Quadratic Formula and the Trig Identities

2. 2cosx^(2)=3sinx+3

In addition to showing your work, so we can see what you did wrong, please confirm that the equations are really

1. \(\displaystyle 2sin^2(x) - 5cos(x) - 4 = 0\)

2. \(\displaystyle 2cos^2(x) = 3sin(x)+3\)

and not what you wrote, which is

1. \(\displaystyle 2sin(x^2) - 5cos(x) - 4 = 0\)

2. \(\displaystyle 2cos(x^2) = 3sin(x)+3\)

Note that if you check your answers in the first equation, you can easily see that they are in fact wrong.

It will also help if you tell us how "it" says you're wrong. If "it" is a computer program, tell us whatever it says specifically; if it is a book, what does it say the right answer is? It could be wrong!