We will learn how to solve different types of problems on trigonometric equation containing one or many trig functions. First we need to solve the trigonometric function (if required) and then solve for the angle value using the trigonometric equation formulas.
1. Solve the equation sec θ  csc θ = 4/3
Solution:
sec θ  csc θ = 4/3
⇒ \(\frac{1}{cos θ}\)  \(\frac{1}{sin θ}\) = 4/3
⇒ \(\frac{sin θ  cos θ}{sin θ cos θ}\) = 4/3
⇒ 3 (sin θ  cos θ) = 4 sin θ cos θ
⇒ 3 (sin θ  cos θ) = 2 sin 2θ
⇒ [3 (sin θ  cos θ)]\(^{2}\) = (2 sin 2θ)\(^{2}\), [Squaring both sides]
⇒ 9 (sin\(^{2}\) θ  2 sin θ cos θ + cos\(^{2}\)
θ) = 4 sin\(^{2}\) 2θ
⇒ 9 (sin\(^{2}\) θ + cos\(^{2}\) θ  2 sin θ cos θ) = 4 sin\(^{2}\) 2θ
⇒ 9 (1  2 sin θ cos θ) = 4 sin\(^{2}\) 2θ
⇒ 4 sin\(^{2}\) 2θ + 9 sin 2θ  9 = 0
⇒ (4 sin 2θ  3)(sin 2θ + 3) = 0
⇒ 4 sin 2θ  3 = 0 or sin 2θ + 3 = 0
⇒ sin 2θ = ¾ or sin 2θ = 3
but sin 2θ = 3 is not possible.
Therefore, sin 2θ = ¾ = sin ∝ (say)
⇒ 2θ = nπ + (1)\(^{n}\) ∝, where, n = 0, ± 1, ± 2, ± 3, ± 4 ..... and sin ∝ = ¾
⇒ θ = \(\frac{nπ}{2}\) + (1)\(^{n}\) \(\frac{∝}{2}\), where, n = 0, ± 1, ± 2, ± 3, ± 4 ..... and sin ∝ = ¾
Therefore, the required solution θ = \(\frac{nπ}{2}\) + (1)\(^{n}\) \(\frac{∝}{2}\), where, n = 0, ± 1, ± 2, ± 3, ± 4 ..... and sin ∝ = ¾
2. Find general solution of the equation cos 4θ = sin 3θ.
Solution:
cos 4θ = sin 3θ
⇒ cos 4θ = cos (\(\frac{π}{2}\)  3θ)
Therefore, 4θ = 2nπ ± (\(\frac{π}{2}\)  3θ)
Therefore, either 4θ = 2nπ + \(\frac{π}{2}\)  3θ Or, 4θ = 2nπ  \(\frac{π}{2}\) + 3x
⇒ 7θ = (4n + 1)\(\frac{π}{2}\) or, θ = (4n  1)\(\frac{π}{2}\)
⇒ θ = (4n + 1)\(\frac{π}{14}\) or, θ = (4n  1)\(\frac{π}{2}\)
Therefore the general solution of the equation cos 4θ = sin 3θ are θ = (4n + 1)\(\frac{π}{14}\)and θ = (4n  1)\(\frac{π}{2}\) , where, n = 0, ±1, ±2, …………………..
`11 and 12 Grade Math
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