Can anyone tell my mistake i am making in the general solution of following equation;
I am getting different answers.
Solve -
4sin2x + 3cos 2x = 5
My answer-
4(2sinxcosx) + 3 ( 2cos2x - 1) = 5
→ 8 sinx cosx + 6 cos2x = 8
→ Divide both sides by cos2x
→ 8 tanx + 6 = 8 (1 + tan2x)
→ 8 tan2x -8 tanx + 2 = 0
→ ( 2tax - 1)2 = 0 → x = n (pi ) + tan-1(1/2) ,where n is an integer
but the answer given in the book is n(pi) + (1/2) tan-1(4/3)
they are actually dividing both sides in original eqn. by 5
and the forming an eqn of type cos (2x - A) where A = cos-1(3/5)
Ultimately we get X = (1/2) cos -1(3/5) which is same as (1/2)tan-1(4/3)
if both answers are same , can anyone prove Tan-1(1/2) = (1/2) tan-1(4/3)
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2 Answers
1tan-1(1/2) = 1/2tan-1(4/3)
this comes from 2tan-1x = tan-1(2x/1-x2)
341a simpler way to solve this particular one is to note that LHS≤5
from a \sin x + b \cos x \le \sqrt{a^2+b^2}
so you immediately get
\cos 2x = \frac{3}{5}; \sin 2x = \frac{4}{5}
