this is a long one
be prepared, it is probably a 13 mark question and is usually breaking down into parts....
(For edexcel C3 By the way)
FOR theta i will be USING "x"
5sin2x + 12cos2x = 11
First thing u need to do is to get the left hand side into a common trig ratio
But since we know that sin(A+B) = sinAcosB + sinBcosA,
we can say that
5sin2x + 12cos2x can be written as Rsin(2x+A)
5sin2x + 12cos2x = Rsin(2x+A)
5sin2x + 12cos2x = Rsin2xcosA + RsinAcos2x
Take the sin2x and cos2x from each side and treat them seperately
therefore,
5sin2x = Rsin2xcosA
5 = RcosA
5/cosA = Rand
12cos2x = RsinAcos2x
12 = RsinASubstituting R from the first part gives us:
12 = 5 sinA/cosA
12/5 = tan A
A = 67.38
R² = 12²+5² (The right angled triangle can be deduced using the earlier formulas in bold)
R = 13
Therefore
5sin2x + 12cos2x = 13sin(2x + 67.38)
Now
13sin(2x+67.
= 11
LET 2x + 67.8 be Y
13sinY = 11
Sin Y = 11/13
Y = 57.796, 122.204, 417.796, 482.24
x = -5.352 ,27.02,174.998,207.22
In the range
x = 27.02,174.998
hope that helped,
if you dont understand something ask away