Please answer the following
#11
Attachments:
Answers
Answered by
2
33= 16(2cos2x+sinx)
⇒33 = 16[2(1- 2sin²x) +sinx]
⇒33 = 16(2 - 4 sin²x + sinx)
⇒33 = 32 - 64sin²x +16sinx
⇒64sin²x -16sinx +1=0
⇒(8sinx -1)² =0
⇒8sinx -1 =0
⇒sinx = 1/8
so, cosx = √[(8)²-(1)²]/8 = √63/8
⇒33 = 16[2(1- 2sin²x) +sinx]
⇒33 = 16(2 - 4 sin²x + sinx)
⇒33 = 32 - 64sin²x +16sinx
⇒64sin²x -16sinx +1=0
⇒(8sinx -1)² =0
⇒8sinx -1 =0
⇒sinx = 1/8
so, cosx = √[(8)²-(1)²]/8 = √63/8
Anonymous:
bhai last step explain kar
sinx = p/h and cosx=b/h...now, b²= h²- p²...now substitute the values
bhai ya konsa class ka formula hai
aur jub b^2 ayaa toh cos^2x hota hai na bhai
iska ans 1/8 hai muja waisa lagta hai
wo pythagores theorem h..jo basic sb padh chuke h...aur we need to know cosx not sinx...aur hmlg ko sinx = 1/8 aaya h...not cosx
Answered by
1
Answer:
33= 16(2cos2x+sinx)
→ 33 = 16[2(1-2sin²x)+sinx] → 33 = 16(2-4 sin²x + sinx)
→ 33 = 32-64sin²x +16sinx
→ 64sin²x -16sinx+1=0
→ (8sinx-1)² = 0
→ 8sinx-1=0
= sinx = 1/8
so, cosx = V[(8)<{1}^)/8 = v63/8
Similar questions