SOLUTION: Find the value of x in the range
0°≤ x ≤ 360°, which satisfy the equation below.
(i) sin3xsinx = 2cos2x + 1
(ii) 1 + cosx = 2sin²x
(iii) 3cotx + tanx - 4 = 0
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-> SOLUTION: Find the value of x in the range
0°≤ x ≤ 360°, which satisfy the equation below.
(i) sin3xsinx = 2cos2x + 1
(ii) 1 + cosx = 2sin²x
(iii) 3cotx + tanx - 4 = 0
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Question 1176499: Find the value of x in the range
0°≤ x ≤ 360°, which satisfy the equation below.
(i) sin3xsinx = 2cos2x + 1
(ii) 1 + cosx = 2sin²x
(iii) 3cotx + tanx - 4 = 0
You can put this solution on YOUR website! ii. 1+ cos x=2-2 cos^2 x, because sin^2x=1-cos^2x
2 cos^2x+ cos x-1=0
(2 cos x-1)(cos x+1)=0
cos x=1/2 (at 60 deg and 300 deg)
cos x=-1 (at 180 deg)
check with 60 deg
1+0.5=2*3/4-1 checks
at 300 deg checks as well
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