where
.
Follow the same process for your second equation.
Sin(2x) - √3/2 **Read as sin(2x) minus square root of three over two** Instructions say to find ALL solutions. However, I believe I am coming up with some negative erroneous solutions.
(2sinx cosx)^2 = (√3/2)^2 **Square both sides**
4sin^2x cos^2x = ¾
4sin^2x (1 – sin^2x) = ¾ **substitute cos2x = 1 – sin2x**
4sin^2x – 4sin^4x = ¾ **distribute**
-4sin^4 + 4sin^2x = ¾ ** re-arrange by descending powers**
-4sin^4 + 4sin^2x – ¾ = 0 **now quadratic in form.**
-4u^2 + 4u – ¾ =0 **let “u” equal sin2**
U = ¾ and ¼
Sin^2x = ¾ and sin^2x = ¼ **take the square root of both sides**
Then sinx = plus or minus √3/2 and sinx = plus or minus √1/2
I believe the quadratic is confusing me due to the plus or minus issue. So I am looking for a way that might be simpler, and quicker without the quadratic.
I have another problem that is very similar:
sin(2x) = 1/2
Thanks in advance.