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c2 trigonometrical identities and simple equestions

cos3q=-1 (q is the degree)
the answers are 60, 180 and 300
so i dont know how to solve this
this what i did, the intervals 3*0 and 3*360=1080
so my values are 180 , 360 , 540, 720, 900, 1080.
cos3q=180 , 360 , 540, 720, 900, 1080
(divide by 3)=60, 120, 180, 300, 360
Reply 1
Avatar for raheem94
raheem94
13
Original post by reb0xx
cos3q=-1 (q is the degree)
the answers are 60, 180 and 300
so i dont know how to solve this
this what i did, the intervals 3*0 and 3*360=1080
so my values are 180 , 360 , 540, 720, 900, 1080.
cos3q=180 , 360 , 540, 720, 900, 1080
(divide by 3)=60, 120, 180, 300, 360


cos3q=−1 \displaystyle cos3q=-1

The interval is, 0≤q≤360 \displaystyle 0\le q \le 360
The interval for '3q' is 0≤3q≤1080 \displaystyle 0\le 3q \le 1080

Solving cos3q=−1 \displaystyle cos3q=-1 will give us 3 solutions. One solution you will get from the calculator the other two solutions can be found by adding 360 and 720 to the calculator value. Now divide the solutions by 3 to get the required values of 'q'.

Below is the graph of y=cos(3q), this shows that it has 3 solution in the interval 0≤q≤360 \displaystyle 0\le q \le 360 .

Original post by reb0xx
cos3q=-1 (q is the degree)
the answers are 60, 180 and 300
so i dont know how to solve this
this what i did, the intervals 3*0 and 3*360=1080
so my values are 180 , 360 , 540, 720, 900, 1080.
cos3q=180 , 360 , 540, 720, 900, 1080
(divide by 3)=60, 120, 180, 300, 360


cos⁡3θ=−1[br][br]cos⁡−1(−1)=180\cos 3\theta = -1 [br][br]\cos ^-1 (-1) = 180

For the next two values, add 360 each time, as the Cosine graph repeats every
360∘[br][br]Wehave180,(180+360),(180+360+360)[br]Whichis;180,540,900[br][br]So,3θ=180,540,900[br]θ=60,180,300360^{\circ} [br][br]We have 180, (180+360), (180+360+360)[br]Which is; 180, 540, 900[br][br]So, 3\theta = 180, 540, 900[br] \theta = 60, 180, 300

Does this help?
(edited 12 years ago)
Reply 3
Avatar for reb0xx
reb0xx
OP
2
Original post by raheem94
cos3q=−1 \displaystyle cos3q=-1

The interval is, 0≤q≤360 \displaystyle 0\le q \le 360
The interval for '3q' is 0≤3q≤1080 \displaystyle 0\le 3q \le 1080

Solving cos3q=−1 \displaystyle cos3q=-1 will give us 3 solutions. One solution you will get from the calculator the other two solutions can be found by adding 360 and 720 to the calculator value. Now divide the solutions by 3 to get the required values of 'q'.

Below is the graph of y=cos(3q), this shows that it has 3 solution in the interval 0≤q≤360 \displaystyle 0\le q \le 360 .



thanks man i solved this
but can you help me with this one
4sinq=tanq
Reply 4
Avatar for raheem94
raheem94
13
Original post by reb0xx
thanks man i solved this
but can you help me with this one
4sinq=tanq


4sinq=tanq \displaystyle 4sinq = tanq

Remember, tanq=sinqcosq \displaystyle tanq=\frac{sinq}{cosq}

4sinq=tanq  ⟹  4sinq=sinqcosq  ⟹  4sinq−sinqcosq=0 \displaystyle 4sinq = tanq \implies 4sinq=\frac{sinq}{cosq} \implies 4sinq - \frac{sinq}{cosq} =0

Now factorise the above expression, e.g. 4x−xy=0  ⟹  x(4−1y)=0 \displaystyle 4x -\frac{x}{y}=0 \implies x\left(4-\frac1{y}\right)=0 This gives x=0 \displaystyle x=0 and
Unparseable latex formula:

\displaystyle 4-\frac1{y}\right = 0

Reply 5
Avatar for reb0xx
reb0xx
OP
2
Original post by raheem94
4sinq=tanq \displaystyle 4sinq = tanq

Remember, tanq=sinqcosq \displaystyle tanq=\frac{sinq}{cosq}

4sinq=tanq  ⟹  4sinq=sinqcosq  ⟹  4sinq−sinqcosq=0 \displaystyle 4sinq = tanq \implies 4sinq=\frac{sinq}{cosq} \implies 4sinq - \frac{sinq}{cosq} =0

Now factorise the above expression, e.g. 4x−xy=0  ⟹  x(4−1y)=0 \displaystyle 4x -\frac{x}{y}=0 \implies x\left(4-\frac1{y}\right)=0 This gives x=0 \displaystyle x=0 and
Unparseable latex formula:

\displaystyle 4-\frac1{y}\right = 0



why did you chose sin to be x ?
Reply 6
Avatar for raheem94
raheem94
13
Original post by reb0xx
why did you chose sin to be x ?


I was giving you an example.
Reply 7
Avatar for reb0xx
reb0xx
OP
2
Original post by raheem94
I was giving you an example.


ok one last question
i can solve this equations using with the quadratic equation formula
this one(http://www.oncalc.com/wp-content/uploads/2011/02/quadratic-equation-calculator.gif)
are we allowed to solve this with a formula or by completing the square instead of factorizing it???
are we allowed to do that in the exams?
Reply 8
Avatar for raheem94
raheem94
13
Original post by reb0xx
ok one last question
i can solve this equations using with the quadratic equation formula
this one(http://www.oncalc.com/wp-content/uploads/2011/02/quadratic-equation-calculator.gif)
are we allowed to solve this with a formula or by completing the square instead of factorizing it???
are we allowed to do that in the exams?


Which question are you talking about?

If the question doesn't specifies a method then you can use any method.

In your previous question, we got, 4sinq−sinqcosq=0 \displaystyle 4sinq - \frac{sinq}{cosq} = 0

Here there is no point of using any formula or completing the square, this can easily be solved by factorisation.

4sinq−sinqcosq=0  ⟹  sinq(4−1cosq)=0 \displaystyle 4sinq - \frac{sinq}{cosq} = 0 \implies sinq\left(4-\frac1{cosq}\right)=0

So we get two equations, sinq=0 \displaystyle sinq=0 and 4−1cosq=0 \displaystyle 4-\frac1{cosq}=0

Do you get it?

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