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Parents Pricing Home NSC Mathematics Trigonometry Given the equation:
sin(x + 60°) + 2cosx = 0
6.1 Show that the equation can be rewritten as tan x = −4 − √3
Given the equation:
sin(x + 60°) + 2cosx = 0
6.1 Show that the equation can be rewritten as tan x = −4 − √3 - NSC Mathematics - Question 6 - 2016 - Paper 2 Question 6
View full question Given the equation:
sin(x + 60°) + 2cosx = 0
6.1 Show that the equation can be rewritten as tan x = −4 − √3.
6.2 Determine the solutions of the equation sin(x + 6... show full transcript
View marking scheme Worked Solution & Example Answer:Given the equation:
sin(x + 60°) + 2cosx = 0
6.1 Show that the equation can be rewritten as tan x = −4 − √3 - NSC Mathematics - Question 6 - 2016 - Paper 2
6.1 Show that the equation can be rewritten as tan x = −4 − √3. Only available for registered users.
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To rewrite the equation, start with:
s i n ( x + 60 ° ) + 2 c o s x = 0 sin(x + 60°) + 2cosx = 0 s in ( x + 60° ) + 2 cos x = 0 Using the angle addition formula for sine, we have:
egin{align*}
sin(x + 60°) &= sinx imes cos60° + cosx imes sin60° \\
sin(x + 60°) &= sinx imes rac{1}{2} + cosx imes rac{ ext{√3}}{2} \\
sin(x + 60°) &= rac{1}{2} sinx + rac{ ext{√3}}{2} cosx
end{align*}
Substituting this into the original equation:
rac{1}{2} sinx + rac{ ext{√3}}{2} cosx + 2cosx = 0
Combine terms:
rac{1}{2} sinx + rac{ ext{√3}}{2} cosx + 2cosx = 0\
\frac{1}{2} sinx + rac{ ext{√3}}{2}cosx + rac{4}{2}cosx = 0 \
\frac{1}{2}sinx + rac{4 + ext{√3}}{2}cosx = 0\
\frac{1}{2}(sinx + (4 + ext{√3})cosx) = 0\
ext{Therefore,} \sin x = -(4 + ext{√3}) cos x
Now, dividing both sides by cosx (assuming it's not zero):
t a n x = − 4 − e x t √ 3 . tan x = -4 - ext{√3}. t an x = − 4 − e x t √3 .
6.2 Determine the solutions of the equation sin(x + 60°) + 2cosx = 0 in the interval −180° ≤ x ≤ 180°. Only available for registered users.
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From the previous part, we have:
t a n x = − 4 − e x t √ 3 . tan x = -4 - ext{√3}. t an x = − 4 − e x t √3 . Using the reference angle:
r e f = 80 ° ref = 80° re f = 80° Then, the solutions in the specified interval are:
x = − 80 ° or x = 180 ° + 80 ° = 99.90 ° x = -80° \
\text{or} \
x = 180° + 80° = 99.90° x = − 80° or x = 180° + 80° = 99.90°
6.3.1 Draw the graph of g(x) = sin(x + 60°) for −120° ≤ x ≤ 240°. Only available for registered users.
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The graph of g(x) can be sketched as follows:
Identify key points for x in the interval −120° to 240°.
Calculate values of g(x) at critical points, specifically at 30°, −60°, and so on.
Sketch the graph smoothly through these points, indicating the oscillatory nature of the sine function.
6.3.2 Determine the values of x in the interval −120° ≤ x ≤ 240° for which sin(x + 60°) + 2cosx > 0. Only available for registered users.
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To find intervals where:
s i n ( x + 60 ° ) + 2 c o s x > 0 , sin(x + 60°) + 2cosx > 0, s in ( x + 60° ) + 2 cos x > 0 , we observe the previously derived function and graphically check where it is above the x-axis.
Through analyzing the graph:
The critical values are when:
x ∈ ( − 80 ° , 10 ° ) or x ∈ ( 80 ° , 180 ° ) e x t , a n d i n t e r v a l s w h e r e ( s i n ( x + 60 ° ) + 2 c o s x ) > 0. x ∈ (-80°, 10°) \
\text{or} \
x ∈ (80°, 180°) ext{, and intervals where }(sin(x + 60°) + 2cosx) > 0. x ∈ ( − 80° , 10° ) or x ∈ ( 80° , 180° ) e x t , an d in t er v a l s w h ere ( s in ( x + 60° ) + 2 cos x ) > 0. Join the NSC students using SimpleStudy...97% of StudentsReport Improved Results
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