cos^2(theta) + 2cos(theta) - 3?

Question

Factor the above expression.

Answer 1

what to do in this

Answer 2

Look, you have to substitute: cos(theta) = t; then your equation becomes:

t^2 + 2t - 3 =0

(I suppose that "=0" is missing in your question; otherwise, it can't be solved, as "n nitant" said, it would be only an expression);

The above equation is an ordinary square equation, and it has two solutions, t1 and t2, given by formula:

t1,2 = [-b ± √(b^2 - 4ac)]/2a, where a = 1 (quote beside t^2), b = 2 (quote beside t), and c = -3 (free quote); therefore,

t1 = 1, t2 = -3;

Obviously, t2 has no sense, because cosine is always between -1 and +1, so we use the first solution; going back to substitution, we have

cos (theta) = 1,
theta = 0 degrees

Hope this was understandable...

Answer 3

Let theta = X

cos² X + 2cos X - 3 = (cos X + 3)(cos X - 1)

Which is required answer as factorising is what has been requested?

But, just in case, if this was an equation = 0:-

cos X = 1

X = 0° , 360°

Answer 4

Let cos theta = x
x^2 + 2x - 3 = 0
( x + 3 )(x - 1 ) = 0
x = -3 or x = 1
cos theta = -3 (rejected) or cos theta = 1
theta = 0° or 360°