Solve for x :
cos x3−1=0\dfrac{x}{3} - 1 = 03x−1=0
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cos x3−1=0\text{cos }\dfrac{x}{3} - 1 = 0cos 3x−1=0
⇒cos x3=1⇒cos x3=cos 0°⇒ \text{cos } \dfrac{x}{3} = 1\\[1em] ⇒ \text{cos } \dfrac{x}{3} = \text{cos 0°}⇒cos 3x=1⇒cos 3x=cos 0°
So,
⇒x3=0°⇒x=3×0°⇒x=0°⇒ \dfrac{x}{3} = 0°\\[1em] ⇒ x = 3 \times 0°\\[1em] ⇒ x = 0°\\[1em]⇒3x=0°⇒x=3×0°⇒x=0°
Hence, x = 0°.
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Find the magnitude of angle A, if :
2 tan 3A cos 3A - tan 3A + 1 = 2 cos 3A
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cos (2x - 30°) = 0