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  • Find the general solution of equation tan3∂=cot2∂
For all angles between -720 and +720

Derick , 3 Years ago
Grade 12th pass
anser 1 Answers
Askiitians Tutor Team

Last Activity: 4 Months ago

To solve the equation \( \tan(3\theta) = \cot(2\theta) \), we can start by rewriting the cotangent function in terms of tangent. Recall that \( \cot(x) = \frac{1}{\tan(x)} \). Therefore, we can express the equation as:

\( \tan(3\theta) = \frac{1}{\tan(2\theta)} \)

By cross-multiplying, we obtain:

\( \tan(3\theta) \tan(2\theta) = 1 \)

This equation implies that the product of the tangents of the two angles equals one. A useful identity here is that \( \tan(A) \tan(B) = 1 \) when \( A + B = \frac{\pi}{2} + n\pi \) for any integer \( n \). Thus, we can set up the following equation:

Setting Up the Equation

From the identity, we have:

\( 3\theta + 2\theta = \frac{\pi}{2} + n\pi \)

Combining the angles gives us:

\( 5\theta = \frac{\pi}{2} + n\pi \)

Solving for \( \theta \)

Now, we can solve for \( \theta \) by dividing both sides by 5:

\( \theta = \frac{\pi}{10} + \frac{n\pi}{5} \)

Next, we need to find the values of \( n \) that keep \( \theta \) within the range of -720 to +720 degrees. First, let's convert the angle from radians to degrees:

\( \frac{\pi}{10} \) radians is equivalent to \( 18^\circ \), and \( \frac{n\pi}{5} \) radians is equivalent to \( 36n^\circ \). So we can rewrite our equation as:

\( \theta = 18^\circ + 36n^\circ \)

Finding Valid Values of \( n \)

Now, we need to determine the integer values of \( n \) such that:

\( -720 \leq 18 + 36n \leq 720 \)

Let's break this down into two inequalities:

  • For the lower bound: \( 18 + 36n \geq -720 \)
  • For the upper bound: \( 18 + 36n \leq 720 \)

Solving the Inequalities

Starting with the lower bound:

\( 36n \geq -720 - 18 \)

\( 36n \geq -738 \)

\( n \geq -\frac{738}{36} \approx -20.5 \)

Thus, \( n \geq -20 \) (since \( n \) must be an integer).

Now for the upper bound:

\( 36n \leq 720 - 18 \)

\( 36n \leq 702 \)

\( n \leq \frac{702}{36} \approx 19.5 \)

Thus, \( n \leq 19 \).

Listing the Integer Values of \( n \)

From our calculations, \( n \) can take any integer value from -20 to 19, inclusive. This gives us a total of:

Values of \( n \): -20, -19, -18, ..., 0, ..., 18, 19

Calculating the Corresponding Angles

Now, we can find the corresponding angles \( \theta \) for each integer \( n \):

  • For \( n = -20 \): \( \theta = 18 - 720 = -702^\circ \)
  • For \( n = -19 \): \( \theta = 18 - 684 = -666^\circ \)
  • For \( n = -18 \): \( \theta = 18 - 648 = -630^\circ \)
  • For \( n = -17 \): \( \theta = 18 - 612 = -594^\circ \)
  • For \( n = -16 \): \( \theta = 18 - 576 = -558^\circ \)
  • For \( n = -15 \): \( \theta = 18 - 540 = -522^\circ \)
  • For \( n = -14 \): \( \theta = 18 - 504 = -486^\circ \)
  • For \( n = -13 \): \( \theta = 18 - 468 = -450^\circ \)
  • For \( n = -12 \): \( \theta = 18 - 432 = -414^\circ \)
  • For \( n = -11 \): \( \theta = 18 - 396 = -378^\circ \)
  • For \( n = -10 \): \( \theta = 18 - 360 = -342^\circ \)
  • For \( n = -9 \): \( \theta = 18 - 324 = -306^\circ \)
  • For \( n = -8 \): \( \theta = 18 - 288 = -270^\circ \)
  • For \( n = -7 \): \( \theta = 18 - 252 = -234^\circ \
  • For \( n = -6 \): \( \theta = 18 - 216 = -198^\circ \
  • For \( n = -5 \): \( \theta = 18 - 180 = -162^\circ \
  • For \( n = -4 \): \( \theta = 18 - 144 = -126^\circ \
  • For \( n = -3 \): \( \theta = 18 - 108 = -90^\circ \
  • For \( n = -2 \): \( \theta = 18 - 72 = -54^\circ \
  • For \( n = -1 \): \( \theta = 18
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