Trigonometeric Equation

help me friends!

Find the general solution of following trigonometric equations-

1) solve cos4x + sin4x = 2 cos ( (2x) + (pi/6)) cos ( (2x) - (pi /6))

2) Solve

sin2x + 2tan2x + ((4/√3) tanx) - sinx + (11/12) = 0

3) 2sin (x + (pi/4)) = tanx + cotx

3 Answers

262
Aditya Bhutra ·

hint for 1st question -
use 2cosAcosB = cos(A+B)+cos(A-B) in rhs
use identity cos 4x = 2cos2(2x)-1 and cos 2x =cos2x-sin2x in rhs
simplify rhs and lhs.
convert terms to sin22x and cos22x then simple stuff.
-

262
Aditya Bhutra ·

hint for q.3

check max. value of lhs
check min value of rhs
compare the two
finish it off :)

11
Devil ·

2)

2\left(tanx+\frac{1}{\sqrt{3}} \right)^2+\left(sinx-\frac{1}{2} \right)^2=0

The general soln. of which is

(2n+1)\pi-\frac{\pi}{6}

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