1. ## General solution

Write general solutions:
4sinacosa = 1
the answer says (npi)/2 + (-1)^n (pi/12)
except when i did it i got a different answer. if anybody could tell me where i went wrong i would be extremely grateful =D
i rooted both sides : 2sinacosa = 1
therefore, sin2a = 1
2a = (pi n) + (-1) ^n (pi/2)
divide by 2 both sides
a = (pi n)/2 + (-1)^n (pi/4)

Solve for 0 b 180
b) secb + tan b = 2cosb
i just realli dont no how to solve this
any help greatly appreciated ! thankyou !

2. Originally Posted by THSKluv
4sinacosa = 1

i rooted both sides : 2sinacosa = 1
you got the wrong root

3. Originally Posted by THSKluv
Write general solutions:
4sinacosa = 1
the answer says (npi)/2 + (-1)^n (pi/12)
except when i did it i got a different answer. if anybody could tell me where i went wrong i would be extremely grateful =D
i rooted both sides : 2sinacosa = 1
therefore, sin2a = 1
2a = (pi n) + (-1) ^n (pi/2)
divide by 2 both sides
a = (pi n)/2 + (-1)^n (pi/4)

Solve for 0 b 180
b) secb + tan b = 2cosb
i just realli dont no how to solve this
any help greatly appreciated ! thankyou !
$4\sin a\cos a=1$

$\Rightarrow 2\sin2a=1$

$\Rightarrow \sin2a=\frac1{2}$

$\Rightarrow \sin2a=\sin\frac\pi{6}$

$\Rightarrow 2a=n\pi+(-1)^n\frac\pi{6}$

$\Rightarrow a=\frac{n\pi}{2}+(-1)^n\frac\pi{12}$

4. oh woops , yea that was stupid, sorry about that, please ignore my solution to the first part
if anyone could help me solve it tho , reaaaly appreciate it thanks !

5. ahh ! thank you so much !
i understand now =DD
THANKS !