Hi Carla,

I want to use the expression that cos²(A) = 1 - sin²(A) so my first though is "what happens if I square both sides?" Squaring gives me
16 sin²(A)cos²(A) + 8 sin(A)cos(A) + 1 = 4(sin²(A) + 2 sin(A)cos(A) + cos²(A))
The sin(A) cos(A) terms cancel and now you can substitute cos²(A) = 1 - sin²(A). This gives a quadratic equation in sin²(A) which factors and should allow you to complete the problem.

If you need more assistance write back,
Harley

Dear Harley,
I am still having difficulties solving this problem: 4sin(A)cos(A) = 2 (sin(A)+cos(A))
After squaring the whole expression and substituting cos² (A) = 1 - sin²(A), I end up with:
16sin²(A) - 16sin⁴(A) = 3
I am stuck there, could you show me the whole solution?
Thanks.
Carla

Well done Carl, that's the equation I got also. Rewrite it as

16 sin⁴(a) - 16 sin²(A) + 3 = 0

Now let x = sin²(A) and substitute into the equation to get

16 x² - 16 x + 3 = 0

Can you factor this quadratic and solve for x?

Harley

Dear Harley,

Here is my solution

4sinAcosA+1=2(sinA+cosA)

Square both sides of the equation:
16sin²Acos² + 8sinAcosA + 1 = 4(sin²A + 2sinAcosA+cos²A)
16sin²Acos² + 1 = 4sin²A + 4cos²A
Substitute cos²A = 1 - sin²A
16sin²A(1 - sin²A) + 1 = 4sin²A + 4(1 - sin²A)
16sin²A - 16sin⁴A + 1=4sin²A + 4 - 4sin²A
16sin²A - 16sin⁴A + 1 = 4
Substitute sin²A = X in order to obtain a quadratic equation:
-16X² + 16X - 3=0
We obtain 2 possible values for sin²A:
sin²A = 3/4 or sin²A = 1/4
sinA = 0.866 or sinA = 0.5
There are 2 possible values for angle A:
60deg and 30deg
Possible further solutions to the equation can be obtained using the following formulas:
sinA = sin (180 - A)
cos A= cos (-A)
therefore, solutions are -60deg, -30deg, +30deg, +60deg, +120deg, +150deg
Substituting each solution in the original equation shows that -30deg and -120deg are not possible (one side of the equation being positive and the other negative)
therefore solutions to the equation in the range -180deg to +180deg are:
-60deg, +30deg, +60deg, +150deg.

I was just wondering if there is a neater way of sifting out the solutions as I have had to substitute each solution in the equation to see if it works...

Cheers

Carla

You really do have to check each of the possible solutions you found to see which are actually solutions to the original problem. I don't know any way to check other than what you did.

Harley